# Projectile motion: dropped and fired bullets land together

> Gravity acts only on the vertical part of a projectile's motion. From that come the parabola, the time of flight, the range, and why 45 degrees goes farthest.

Series: Physics (https://www.monocurl.com/monocurl-essays/physics/)
Canonical: https://www.monocurl.com/monocurl-essays/physics/projectile-motion/
Keywords: projectile motion, range formula, time of flight, maximum height, trajectory equation, 45 degree launch angle, independence of horizontal and vertical motion, complementary angles

[Figure 1, animation: Two balls leave a ledge 4.9 m high at the same instant, shown at half speed: one is dropped and one is fired sideways at 6 m/s. A ring marks each ball every tenth of a second, and the rings stay level with each other all the way down.]

A rifle is held level, 1.5 m above flat ground, and fired. At the instant the bullet leaves the barrel, a second bullet is dropped from the same height. One of them is moving at 400 m/s and will come down two hundred meters away; the other falls at your feet. Which hits the ground first?

**Pause and try.** Commit to an answer and a reason before reading on: the dropped bullet, the fired bullet, or neither.

**Answer.** Neither. With air resistance ignored they land at the same instant, about 0.55 s after release. Most people pick the dropped bullet, feeling that the fired one's speed should hold it up for a while. The animation above runs the same experiment with slower balls, and the rest of the lesson explains why forward speed gives no protection against falling.

## What the strobe picture shows

In the animation the two balls leave a ledge 4.9 m high. Read the heights of the rings. After 0.2 s both balls have fallen 0.20 m; after 0.4 s, 0.78 m; after 0.6 s, 1.76 m; after 1.0 s, 4.9 m, and both are on the ground. These are the distances $\tfrac12 g t^2 = 4.9t^2$ that a body dropped from rest falls in $t$ seconds, where $g = 9.8$ m/s² is the acceleration of free fall. The fired ball falls exactly as if it had been dropped.

Now read the fired ball's horizontal positions: 1.2 m, 2.4 m, 3.6 m, and 6 m at landing. It covers 0.6 m in every tenth of a second, which is a steady 6 m/s, exactly as if gravity were absent.

So the fired ball seems to be doing two simple things at once: moving sideways at constant velocity, and falling like a dropped ball. Neither motion takes any notice of the other. That is a strong claim, and a picture of one experiment does not prove it. Newton's second law does.

## Why the two motions are independent

The [second law](@physics/newtons-second-law) says that the net force on an object equals its mass times its acceleration, and that this holds separately along each of two perpendicular axes: $\sum F_x = ma_x$ and $\sum F_y = ma_y$. Take $x$ horizontal, in the direction the ball is fired, and $y$ vertically upward.

Once the ball is in the air, and if we neglect air resistance, exactly one force acts on it: its weight, of magnitude $mg$, straight down. Whatever threw the ball stopped pushing when the ball left it. The weight has no horizontal component, so

$$
0 = m a_x \quad\text{and}\quad -mg = m a_y, \qquad\text{that is,}\qquad a_x = 0, \quad a_y = -g.
$$

Look at what each equation contains. The vertical one says the vertical velocity changes at the rate $-g$, and it mentions neither the horizontal position nor the horizontal velocity. A ball moving sideways at 6 m/s, a bullet moving at 400 m/s and a ball not moving sideways at all obey the same vertical equation, and if they start with the same height and the same vertical velocity (zero), they have the same height at every later moment. The horizontal equation says the horizontal velocity never changes, and it does not mention the vertical motion.

**Key idea.** With gravity as the only force, a projectile's horizontal motion has constant velocity and its vertical motion has constant acceleration $-g$, and neither affects the other. The only thing the two motions share is the clock.

The independence is a consequence of this particular force, and it is worth knowing how it could fail. If the force on the ball had a horizontal part that depended on the vertical motion, the two equations would be tied together. Air resistance does exactly that, as we will see at the end.

[Figure 2, animation: A ball kicked at 20 m/s, 60 degrees above the ground, in real time. Its shadow on the ground (blue) moves at a constant 10 m/s; its shadow on the wall (orange) rises and falls like a ball thrown straight up at 17.3 m/s. A dot is left every half second.]

## The equations of a projectile

To use the idea we need the two one-dimensional motions as formulas. Both are motions with constant acceleration, and for those the [earlier lesson on motion graphs](@physics/motion-graphs) derived, from the area under a velocity graph, that an object starting at position $x_0$ with velocity $v_0$ and constant acceleration $a$ has

$$
v = v_0 + at, \qquad x = x_0 + v_0 t + \tfrac12 a t^2.
$$

We also need the starting velocity along each axis. A launch at speed $v_0$ and angle $\theta$ above the horizontal is an arrow of length $v_0$ tilted at $\theta$. It is the hypotenuse of a right triangle whose horizontal side is $v_0\cos\theta$ and whose vertical side is $v_0\sin\theta$; those two sides are the horizontal and vertical components of the launch velocity.

**Theorem.** (Projectile motion.) A point object is launched at time $t = 0$ from the origin with speed $v_0$ at an angle $\theta$ above the horizontal. Take $x$ horizontal in the direction of launch and $y$ vertically up. Assume the frame is inertial and that the only force on the object is its weight $mg$, with $g$ constant (no air resistance, and a region small enough that the ground is flat and gravity uniform). Then while it is in flight, its velocity components (in m/s) and coordinates (in m) at time $t$ (in s) are

$$
v_x = v_0\cos\theta, \qquad v_y = v_0\sin\theta - g t,
$$

$$
x = (v_0\cos\theta)\,t, \qquad y = (v_0\sin\theta)\,t - \tfrac12 g t^2.
$$

**Proof.** The second law along each axis gives $a_x = 0$ and $a_y = -g$, as shown above; both are constant. Apply the constant-acceleration equations to the $x$ coordinate with starting position 0, starting velocity $v_0\cos\theta$ and acceleration 0: $v_x = v_0\cos\theta$ and $x = (v_0\cos\theta)t$. Apply them to the $y$ coordinate with starting position 0, starting velocity $v_0\sin\theta$ and acceleration $-g$: $v_y = v_0\sin\theta - gt$ and $y = (v_0\sin\theta)t - \tfrac12 g t^2$.

The fired ball of the opening is the case $\theta = 0$, launched from height 4.9 m: $x = 6t$ and the height is $4.9 - 4.9t^2$, which is zero at $t = 1$ s whatever the 6 is replaced by.

Here is a more typical case, the one in the last animation: a ball kicked from the ground at $v_0 = 20$ m/s and $\theta = 60°$. Its velocity components at launch are $20\cos 60° = 10$ m/s and $20\sin 60° = 17.3$ m/s. The vertical motion decides how long it stays up. It rises until $v_y = 0$, which happens when $17.3 - 9.8t = 0$, at $t = 1.77$ s, and by then it has climbed to $y = 17.3(1.77) - 4.9(1.77)^2 = 15.3$ m. It returns to the ground when $y = 0$ again: $t\,(17.3 - 4.9t) = 0$ gives $t = 3.53$ s, twice the time to the top. All that while it has been moving sideways at 10 m/s, so it lands $10 \times 3.53 = 35.3$ m away.

**Pause and try.** Repeat the calculation for the same kick speed at a lower angle: $v_0 = 20$ m/s and $\theta = 30°$. Find the time of flight, the greatest height and the landing distance. Compare the landing distance with the $60°$ kick.

**Answer.** The components swap: $v_x = 20\cos 30° = 17.3$ m/s and $v_y = 20\sin 30° = 10$ m/s at launch. Top of the flight: $10 - 9.8t = 0$, so $t = 1.02$ s, at height $10(1.02) - 4.9(1.02)^2 = 5.1$ m. Landing: $t = 2.04$ s, at $x = 17.3 \times 2.04 = 35.3$ m. The low kick is in the air for less time and rises a third as high, and it lands in the same place as the high one. The reason comes out of the general formulas below.

## The path is a parabola

There is a second way to read the equation $y = (v_0\sin\theta)t - \tfrac12 gt^2$, and it makes the shape of the path visible. The first term alone is where the ball would be if gravity were switched off: moving along its line of aim at a steady $v_0$. The second term is the distance a dropped ball falls in the same time.

[Figure 3, steps: The 30 degree kick at 20 m/s. Without gravity the ball would move along its line of aim; with gravity it is below that line by the distance a dropped ball falls in the same time.]

  - step: With gravity switched off the ball would follow its line of aim at a steady 20 m/s, covering 10 m every half second.
  - step: Gravity puts it below that line by the distance a dropped ball falls in the same time: 1.2 m, 4.9 m, 11.0 m and 19.6 m.
  - step: The actual positions lie on one smooth curve, the path of the ball. It lands just after 2 s.
To get the curve as an equation in $x$ and $y$ alone, eliminate the clock. From the horizontal motion $t = x/(v_0\cos\theta)$, which is allowed when $\cos\theta \neq 0$, that is, when the launch is not straight up. Substituting into $y$:

$$
y = (v_0\sin\theta)\frac{x}{v_0\cos\theta} - \frac{g}{2}\left(\frac{x}{v_0\cos\theta}\right)^2 = x\tan\theta - \frac{g}{2v_0^2\cos^2\theta}\,x^2.
$$

This has the form $y = bx - cx^2$ with $c > 0$, the equation of a parabola opening downward. The first term, $x\tan\theta$, is the line of aim, and the second is the fall below it.

[Figure 4, image: The 60 degree kick at four moments a quarter of the flight apart. The horizontal part of the velocity (blue) is the same every time; the vertical part (orange) shrinks, vanishes at the top and reverses.]

## Time of flight, height and range

The two worked kicks followed the same three steps, so we can run those steps once with letters. The results hold for a projectile that lands at the height it was launched from.

**Theorem.** (Level-ground formulas.) Under the assumptions of the projectile theorem, with $0° < \theta < 90°$, suppose the projectile lands at its launch height. Then

- its time of flight is $T = \dfrac{2v_0\sin\theta}{g}$;
- its greatest height is $H = \dfrac{v_0^2\sin^2\theta}{2g}$, reached at time $T/2$;
- its range (horizontal distance to the landing point) is $R = \dfrac{2v_0^2\sin\theta\cos\theta}{g}$;
- for a fixed launch speed $v_0$, the range is greatest at $\theta = 45°$, where $R = v_0^2/g$, and the launch angles $\theta$ and $90° - \theta$ give equal ranges.

**Proof.** (1) The projectile is at launch height when $y = 0$: $t\,(v_0\sin\theta - \tfrac12 g t) = 0$. The root $t = 0$ is the launch, and the other is $T = 2v_0\sin\theta/g$.

(2) While $v_y > 0$ the height is increasing and while $v_y < 0$ it is decreasing, so the greatest height occurs when $v_y = 0$: $t = v_0\sin\theta/g = T/2$. Then

$$
H = (v_0\sin\theta)\frac{v_0\sin\theta}{g} - \frac{g}{2}\,\frac{v_0^2\sin^2\theta}{g^2} = \frac{v_0^2\sin^2\theta}{2g}.
$$

(3) The horizontal velocity is constant, so $R = (v_0\cos\theta)\,T = 2v_0^2\sin\theta\cos\theta/g$.

(4) We must find the largest value of the product $\sin\theta\cos\theta$. For any angle, $(\sin\theta - \cos\theta)^2 \ge 0$. Expanding and using $\sin^2\theta + \cos^2\theta = 1$ gives $1 - 2\sin\theta\cos\theta \ge 0$, so $\sin\theta\cos\theta \le \tfrac12$, with equality exactly when $\sin\theta = \cos\theta$. Between $0°$ and $90°$ that happens only at $\theta = 45°$. So $R \le v_0^2/g$, with equality only at $45°$. For the second claim, $\sin(90° - \theta) = \cos\theta$ and $\cos(90° - \theta) = \sin\theta$, so replacing $\theta$ by $90° - \theta$ swaps the two factors of the product and leaves $R$ unchanged.

Since $2\sin\theta\cos\theta = \sin 2\theta$ (the double-angle identity), the range is often written $R = v_0^2\sin 2\theta/g$. A units check: $v_0^2/g$ is (m²/s²)/(m/s²) = m, and the trigonometric factors are pure numbers, so $R$ and $H$ are lengths; $v_0/g$ is (m/s)/(m/s²) = s, so $T$ is a time.

**Note.** If you know calculus, part 4 is a one-line [optimization](@calculus/optimization): the derivative of $\sin 2\theta$ is $2\cos 2\theta$, which is zero at $\theta = 45°$.

The proof of part 4 also says *why* $60°$ and $30°$ tied. Range is horizontal speed times time aloft, and time aloft is proportional to vertical launch speed. The $60°$ kick has vertical speed 17.3 and horizontal speed 10; the $30°$ kick has them the other way round. The product is the same. At $45°$ the two components are equal, and a product of two numbers whose squares have a fixed sum is largest when the numbers are equal.

**Pause and try.** A ball is kicked at 20 m/s at angles of $15°$, $30°$, $45°$, $60°$ and $75°$. Before watching the next animation, rank the five landing distances, and say which kick is in the air longest.

**Answer.** $45°$ lands farthest, at $20^2/9.8 = 40.8$ m. $30°$ and $60°$ tie at 35.3 m, and $15°$ and $75°$ tie at $40.8 \times \sin 30° = 20.4$ m. The $75°$ kick is in the air longest, 3.94 s, because time of flight depends only on the vertical launch speed.

[Figure 5, animation: Five balls kicked together at 20 m/s, at angles from 15 to 75 degrees, in real time. The 45 degree ball lands farthest, and the pairs 30 and 60 degrees, 15 and 75 degrees, come down at the same points at different times.]

[Figure 6, image: Range against launch angle for a launch speed of 20 m/s on level ground. The curve is symmetric about 45 degrees, where it peaks at 40.8 m; angles equally far from 45 degrees on either side give the same range.]

## Worked examples

**Worked example.** **The two bullets.** A bullet leaves a level rifle barrel 1.5 m above flat ground at 400 m/s. Neglecting air resistance, when and where does it land?

The vertical motion alone sets the time. Take the origin on the ground below the muzzle, with $y$ up. The bullet starts at $y_0 = 1.5$ m with zero vertical velocity, so $y = 1.5 - 4.9t^2$. Setting $y = 0$: $t = \sqrt{1.5/4.9} = 0.553$ s. The 400 m/s never entered, so a dropped bullet takes the same 0.553 s. Horizontally, $x = 400 \times 0.553 = 221$ m.

*Check.* Units: $\sqrt{\text{m}/(\text{m/s}^2)} = \text{s}$. A fall of 1.5 m in about half a second matches everyday experience of dropping something from chest height.

**Worked example.** **Hitting a target.** A ball is kicked from the ground at 20 m/s. At what angle should it be launched to land 30 m away on level ground?

We know $R$ and $v_0$ and want $\theta$, so we use the range formula in its double-angle form: $\sin 2\theta = gR/v_0^2 = 9.8 \times 30/400 = 0.735$. Between $0°$ and $180°$ two angles have this sine: $2\theta = 47.3°$ and $2\theta = 180° - 47.3° = 132.7°$. So $\theta = 23.7°$ or $\theta = 66.3°$, a low shot and a high shot. They add to $90°$, as part 4 of the theorem requires.

The two differ in time of flight: $T = 2(20)\sin 23.7°/9.8 = 1.64$ s for the low shot and $2(20)\sin 66.3°/9.8 = 3.74$ s for the high one. *Check:* horizontal speed times time must be 30 m. Low shot: $20\cos 23.7° \times 1.64 = 30.0$ m.

**Worked example.** **A throw from a cliff.** A stone is thrown from the edge of a cliff 20 m above the sea, at 15 m/s and $40°$ above the horizontal. When and where does it hit the water, and how fast is it moving?

The landing point is below the launch point, so the level-ground formulas do not apply, and we go back to the component equations. Origin at the base of the cliff, $y$ up. Launch components: $v_x = 15\cos 40° = 11.49$ m/s and $v_{y0} = 15\sin 40° = 9.64$ m/s. The height is $y = 20 + 9.64t - 4.9t^2$. Setting $y = 0$ and using the quadratic formula,

$$
t = \frac{9.64 + \sqrt{9.64^2 + 4(4.9)(20)}}{9.8} = \frac{9.64 + 22.02}{9.8} = 3.23 \text{ s},
$$

where we kept the positive root. Then $x = 11.49 \times 3.23 = 37.1$ m from the base of the cliff. At impact $v_y = 9.64 - 9.8 \times 3.23 = -22.0$ m/s and $v_x$ is still 11.49 m/s, so the speed is $\sqrt{11.49^2 + 22.0^2} = 24.8$ m/s.

*Check.* The vertical motion satisfies $v_y^2 = v_{y0}^2 + 2(-g)(\Delta y)$ with $\Delta y = -20$ m, so $v_y^2 = 93.0 + 392 = 485.0$ and $|v_y| = 22.0$ m/s.

[Figure 7, animation: The stone thrown from the 20 m cliff at 15 m/s and 40 degrees, in real time. Its speed drops to 11.5 m/s at the top of the path, where only the horizontal velocity is left, and reaches 24.8 m/s at the water.]

**Common mistake.** **"At the top of its path the ball stops."** A ball thrown straight up does stop for an instant at the top. A projectile launched at an angle does not: only its vertical velocity is zero there, and it is still moving horizontally at $v_0\cos\theta$. The stone above never moves slower than 11.5 m/s. Its acceleration at the top is $g$ downward, the same as everywhere else on the path.

**Common mistake.** **"The force of the throw carries it forward."** It is tempting to draw a forward force on a ball in flight, because the ball is moving forward. The hand stopped pushing at release, and motion at constant velocity needs no force. After release the free-body diagram contains the weight and nothing else.

## What air resistance changes

Every result above assumed that gravity is the only force. For a shot put, or a ball tossed across a room, that is accurate to a percent or so. For a fast, light object it is not. Air resistance is a force directed against the velocity, and for a ball at sporting speeds its size grows roughly as the square of the speed.

[Figure 8, image: A baseball hit at 45 m/s and 45 degrees: the parabola it would follow with no air, and a path computed with air resistance proportional to the square of the speed, for a ball whose terminal speed is 40 m/s. The table above the plot gives the range and the peak height of each path, and a dot marks each peak.]

Four things change, and each can be read off the figure or the force.

- **The range and the height shrink.** In this model the baseball's range drops from 207 m to about 111 m. Real fly balls hit this hard travel about 120 m.
- **The path is no longer symmetric.** The ball loses horizontal velocity all the way along, so it comes down more steeply than it went up, and the highest point is past the middle of the range.
- **The best angle is below $45°$.** A flatter launch spends less time fighting the air. For this model the greatest range is at about $40°$.
- **The two motions are no longer independent.** The drag force points against the velocity and its size depends on the total speed, so the vertical equation now contains the horizontal velocity. A fast horizontal bullet feels a larger vertical drag while it falls than a dropped one does, and it lands slightly later. The dropped and fired bullets land together only in the model without air.

The model has other boundaries. The ground was taken to be flat and $g$ constant; for an artillery shell that travels tens of kilometers, or a rocket, the Earth's curvature and the weakening of gravity with height both matter. And if the launch and landing heights differ, as in the cliff example, the component equations still hold but the level-ground formulas, and with them the $45°$ rule, do not.

**Check yourself.** A soccer ball is kicked at an angle and air resistance is negligible. At the highest point of its path, which statement is true?
Answer: Its velocity is horizontal and its acceleration is $g$ downward $v_y = 0$ at the top, so the velocity is $v_0\cos\theta$ along the horizontal. The only force is the weight, so the acceleration is $g$ downward, as at every point of the flight.

**Recap.** - With gravity the only force, $a_x = 0$ and $a_y = -g$. The horizontal motion has constant velocity, the vertical motion is free fall, and they share only the time.
- From the origin: $x = (v_0\cos\theta)t$, $y = (v_0\sin\theta)t - \tfrac12 gt^2$, $v_x = v_0\cos\theta$, $v_y = v_0\sin\theta - gt$.
- The path is the parabola $y = x\tan\theta - \dfrac{g}{2v_0^2\cos^2\theta}x^2$: the line of aim minus a free fall.
- Landing at launch height: $T = 2v_0\sin\theta/g$, $H = v_0^2\sin^2\theta/(2g)$, $R = v_0^2\sin 2\theta/g$.
- On level ground the range is greatest at $45°$, and $\theta$ and $90° - \theta$ give the same range.
- Air resistance shortens and steepens the path, lowers the best angle, and couples the two motions.

In the cliff example the landing speed could have been found without the angle at all. The next lesson, on [conservation of energy](@physics/energy-conservation), explains why.

## Problems

Use $g = 9.8$ m/s² and neglect air resistance unless a problem says otherwise.

**Problem 1.** A ball rolls off the edge of a table 1.2 m high at 2.5 m/s. How long is it in the air, and how far from the foot of the table does it land?

**Solution.** Origin on the floor below the edge, $y$ up, $x$ along the ball's direction. Vertically the ball starts at 1.2 m with $v_y = 0$: $0 = 1.2 - 4.9t^2$, so $t = \sqrt{1.2/4.9} = 0.495$ s. Horizontally $x = 2.5 \times 0.495 = 1.24$ m.

**Problem 2.** A ball is kicked from level ground at 18 m/s and $35°$. Find its time of flight, range and greatest height.

**Solution.** $v_0\sin\theta = 18 \times 0.5736 = 10.32$ m/s and $v_0\cos\theta = 18 \times 0.8192 = 14.74$ m/s. $T = 2(10.32)/9.8 = 2.11$ s. $R = 14.74 \times 2.11 = 31.1$ m (the formula $18^2\sin 70°/9.8$ gives the same). $H = 10.32^2/(2 \times 9.8) = 5.44$ m.

**Problem 3.** For the kick in the previous problem, find the ball's speed at the top of its path, and its speed and direction of motion at the instant it lands.

**Solution.** At the top $v_y = 0$, so the speed is $v_x = 14.74$ m/s. At landing $t = T$, and $v_y = v_0\sin\theta - gT = v_0\sin\theta - 2v_0\sin\theta = -v_0\sin\theta = -10.32$ m/s, while $v_x = 14.74$ m/s. The speed is $\sqrt{14.74^2 + 10.32^2} = 18$ m/s, the launch speed, and the direction is $35°$ below the horizontal. On level ground a projectile lands with its launch speed, at its launch angle mirrored downward.

**Problem 4.** A stone thrown horizontally from the top of a cliff hits the sea 2.0 s later, 24 m from the base of the cliff. Find the speed of the throw, the height of the cliff, and the stone's speed and direction at impact.

**Solution.** Horizontal: $24 = v_0 \times 2.0$, so $v_0 = 12$ m/s. Vertical: the stone falls $\tfrac12(9.8)(2.0)^2 = 19.6$ m, the height of the cliff. At impact $v_x = 12$ m/s and $v_y = -9.8 \times 2.0 = -19.6$ m/s, so the speed is $\sqrt{12^2 + 19.6^2} = 23.0$ m/s, directed at $\tan^{-1}(19.6/12) = 58.5°$ below the horizontal.

**Problem 5.** Show that for a projectile on level ground, $H/R = \tfrac14\tan\theta$. What is the greatest height of a $45°$ launch as a fraction of its range? At what launch angle does a projectile rise exactly as high as it travels?

**Solution.** $\dfrac{H}{R} = \dfrac{v_0^2\sin^2\theta}{2g} \cdot \dfrac{g}{2v_0^2\sin\theta\cos\theta} = \dfrac{\sin\theta}{4\cos\theta} = \tfrac14\tan\theta$. At $45°$, $\tan\theta = 1$ and $H = R/4$: the 20 m/s kick at $45°$ has $R = 40.8$ m and $H = 10.2$ m. $H = R$ requires $\tan\theta = 4$, that is, $\theta = 76.0°$. The speed and $g$ cancel, so these statements hold for any launch speed, on any planet.

**Problem 6.** A zookeeper aims a tranquilizer dart straight at a monkey hanging from a branch. At the instant the dart is fired, the monkey lets go and drops. Show that the dart hits the monkey whatever the dart's speed, provided the dart reaches the monkey's vertical line before either hits the ground.

**Solution.** Use the "line of aim minus a free fall" description. Let $P$ be the monkey's starting point. With gravity off, the dart would travel along the straight line of aim and reach $P$ at some time $t^*$ (equal to the horizontal distance to the monkey divided by $v_0\cos\theta$). With gravity on, the dart at time $t^*$ is below $P$ by $\tfrac12 g t^{*2}$, directly under it, since gravity does not change the horizontal motion. The monkey started at $P$ at rest and has been falling for the same time $t^*$, so it is also $\tfrac12 g t^{*2}$ below $P$. Dart and monkey are at the same point at the same time. A faster dart only makes $t^*$ smaller, so they meet higher up; the argument fails only if $\tfrac12 g t^{*2}$ exceeds the monkey's starting height, in which case both reach the ground first.

**Problem 7.** A stone is thrown at speed $v_0$ from a height $h$ above level ground. (a) Prove that its speed when it reaches the ground is $\sqrt{v_0^2 + 2gh}$, whatever the launch angle. (b) Check this against the cliff example in the lesson.

**Solution.** (a) Let the launch angle be $\theta$ (positive, zero or negative). The horizontal velocity stays $v_x = v_0\cos\theta$. The vertical motion has constant acceleration $-g$ and vertical displacement $-h$, and for constant acceleration $v^2 = v_{\text{start}}^2 + 2a\,\Delta y$, so $v_y^2 = v_0^2\sin^2\theta + 2gh$. The landing speed satisfies

$$
v^2 = v_x^2 + v_y^2 = v_0^2\cos^2\theta + v_0^2\sin^2\theta + 2gh = v_0^2 + 2gh,
$$

using $\sin^2\theta + \cos^2\theta = 1$. The angle has dropped out. (b) $\sqrt{15^2 + 2(9.8)(20)} = \sqrt{617} = 24.8$ m/s, as found there.

**Problem 8.** From the 20 m cliff of the lesson, with the same throwing speed of 15 m/s, compare the horizontal distances reached by throws at $45°$ and at $30°$. What does the result say about the $45°$ rule?

**Solution.** Use $y = 20 + (v_0\sin\theta)t - 4.9t^2 = 0$ for each. *At $45°$:* both components are $15 \times 0.7071 = 10.61$ m/s. $t = \dfrac{10.61 + \sqrt{10.61^2 + 392}}{9.8} = \dfrac{10.61 + 22.46}{9.8} = 3.37$ s, and $x = 10.61 \times 3.37 = 35.8$ m. *At $30°$:* $v_{y0} = 7.5$ m/s and $v_x = 12.99$ m/s. $t = \dfrac{7.5 + \sqrt{7.5^2 + 392}}{9.8} = \dfrac{7.5 + 21.17}{9.8} = 2.93$ s, and $x = 12.99 \times 2.93 = 38.0$ m. The $30°$ throw goes farther. When the landing point is below the launch point the stone has extra time in the air however it is thrown, so horizontal speed is worth more than it is on level ground, and the best angle falls below $45°$. Part 4 of the theorem assumed equal launch and landing heights, and that hypothesis cannot be dropped.