Angled Projectiles

The video lesson for this topic is coming soon.

Essential Question

How far and how high does a launched object go?

Punts, jump shots, and golf drives all leave the ground at an angle, rise, and come back down. By the end of this lesson, you should be able to split a launch velocity into components and find a projectile’s hang time, maximum height, and range.

Launching at an Angle

Split v_i into components
v_x = v_i cos θ
v_iy = v_i sin θ

A ball with a purple launch velocity arrow v i at angle theta above the horizontal. A dashed red horizontal component is labeled v x = v i cos theta, and a dashed green vertical component is labeled v i y = v i sin theta.

When an object is launched at an angle θ above the horizontal with speed v_i, the first step is always to split the launch velocity into components, just like in the Vector Components lesson. The horizontal component is v_x = v_i cos θ, and the initial vertical component is v_iy = v_i sin θ. From then on, we treat the horizontal and vertical motions separately.

Velocity Along the Path

v_x never changes
v_y shrinks going up
v_y grows coming down

A symmetric parabola with seven evenly timed ball positions. Every ball has a blue horizontal arrow of the same length. Red vertical arrows point up and shrink on the way up, there is no red arrow at the top where v y = 0, and red arrows point down and grow on the way down.

Here is the velocity of an angled projectile at equal time steps. The blue horizontal arrows never change, because no horizontal force acts. The red vertical arrows point up and shrink on the way up, because gravity takes away 9.8 m/s of upward velocity every second. At the top the vertical velocity is zero, and on the way down the red arrows point down and grow. The vertical motion is exactly the tossed-ball motion from the Free Fall lesson.

At the Top

v_y = 0
v_x is NOT zero
a is still 9.8 m/s² down

A close-up of the top of a projectile's parabolic path. At the peak, the ball has a blue horizontal arrow labeled v x, a label v y = 0, and a green arrow pointing straight down labeled a = −9.8 m/s squared. Caption: at the top, the velocity is all horizontal.

At the highest point, the projectile stops rising for an instant, so v_y = 0. But unlike a ball tossed straight up, it is still moving: its horizontal velocity v_x has not changed. So at the top, the speed is not zero, it is just v_x. And the acceleration is still 9.8 m/s² downward, because gravity never switches off. The fact that v_y = 0 at the top is the key to finding both hang time and maximum height.

The Path Is Symmetric

Time up = time down
Same height, same speed
Level ground: lands at launch speed

A symmetric parabola with a dashed vertical line through the peak. Two balls at the same height on opposite sides have purple velocity arrows of equal length, one pointing up and right, the other down and right. Below the ground, double arrows mark t up from launch to the middle and t down from the middle to landing. Caption: the path is a mirror image.

On level ground, a projectile’s path is perfectly symmetric about its highest point. It takes the same time to go up as to come down. At any height, the projectile has the same speed on the way down as it had on the way up, just angled downward instead of upward. So it lands with the same speed it was launched with. This symmetry is a huge shortcut: find the time to reach the top, then double it to get the total time.

Hang Time

Use the vertical motion only
Time to top: v_y goes to 0
Total time = 2 × time to top

A graph of vertical velocity versus time. A straight line starts at plus v i y, crosses zero at t up, where a red dot marks the top, and ends at minus v i y at t total. Caption: t total equals 2 times t up.

Hang time comes from the vertical motion alone. The vertical velocity starts at +v_iy and drops by 9.8 m/s every second until it reaches 0 at the top. Using v_fy = v_iy + at with v_fy = 0 gives the time to the top: t_up = v_iy / 9.8 m/s². On level ground the trip down takes the same time, so the total time in the air is 2 × t_up.

Maximum Height

Use the vertical motion only
At the top, v_fy = 0
v_fy² = v_iy² + 2aΔy

A dashed parabola rising from the ground to a peak and back down. At the peak, a ball is labeled v y = 0. A double arrow from the peak straight down to the ground is labeled H. Caption: stop at the top, where v f y = 0.

The maximum height also comes from the vertical motion. Stop the problem at the top, where the final vertical velocity is zero. Then use v_fy² = v_iy² + 2aΔy with v_fy = 0 and a = −9.8 m/s², and solve for Δy. You can also find the time to the top first and use Δy = v_iyt + ½at². Either way, only the vertical component v_iy matters, not the full launch speed.

Range

Range = horizontal distance
Δx = v_x × t_total
45° gives the most range

Three dashed parabolas launched from the same spot at the same speed. The 60 degree path (blue) goes highest, the 45 degree path (green) goes farthest, and the 30 degree path (red) is lowest. The 30 and 60 degree paths land at the same spot. Caption: same speed, 30 degrees and 60 degrees land together.

The range is the horizontal distance a projectile travels before it lands. Because the horizontal velocity is constant, range = v_x × t_total. A steep launch spends a long time in the air but moves forward slowly, and a shallow launch moves forward fast but lands quickly. On level ground with no air resistance, 45° is the best compromise and gives the maximum range. Angles that add up to 90°, like 30° and 60°, give the same range.

Worked Examples

Try each problem on your own first, then check your work against the solution.

Example 1: Launch Components

A soccer ball is kicked at 19.6 m/s at 30.0° above the ground. Find its horizontal velocity and its initial vertical velocity.

A soccer ball with a purple launch arrow of 19.6 m/s at 30.0 degrees above the ground. A dashed red horizontal component is labeled v x = ? and a dashed green vertical component is labeled v i y = ?.

Knowns: v_i = 19.6 m/s, θ = 30.0°
Horizontal: v_x = v_i cos θ = (19.6 m/s)(cos 30.0°) = (19.6 m/s)(0.866)
Vertical: v_iy = v_i sin θ = (19.6 m/s)(sin 30.0°) = (19.6 m/s)(0.500)
Answer: v_x = 17.0 m/s and v_iy = 9.80 m/s (upward)

These two numbers are all you need for every other question about this kick: v_x stays 17.0 m/s the whole flight, and v_iy = 9.80 m/s starts the free-fall motion.

Example 2: Hang Time

A football is kicked from the ground at 25.0 m/s at 36.9° above the horizontal. How long is it in the air before it lands on the field?

A football's dashed parabolic path from the ground back to the ground. A purple launch arrow is labeled 25.0 m/s at 36.9 degrees, and a double arrow along the ground is labeled t = ?.

Vertical component: v_iy = v_i sin θ = (25.0 m/s)(sin 36.9°) = 15.0 m/s
Time to the top: At the top v_fy = 0. From v_fy = v_iy + at: t_up = (0 − 15.0 m/s) / (−9.8 m/s²) = 1.53 s
Total time: The path is symmetric, so t_total = 2 × 1.53 s
Answer: t = 3.06 s

Example 3: Maximum Height

A water balloon is launched at 14.0 m/s at 60.0° above the horizontal. What is its maximum height above the launch point?

A water balloon's dashed parabolic path. A purple launch arrow is labeled 14.0 m/s at 60.0 degrees. A double arrow from the peak down to the launch level is labeled H = ?.

Vertical component: v_iy = (14.0 m/s)(sin 60.0°) = 12.1 m/s
Knowns (up to the top): v_iy = 12.1 m/s, v_fy = 0, a = −9.8 m/s², Δy = ?
Equation: v_fy² = v_iy² + 2aΔy, so Δy = (v_fy² − v_iy²) / 2a
Substitute: Δy = (0 − (12.1 m/s)²) / 2(−9.8 m/s²) = (−147 m²/s²) / (−19.6 m/s²)
Answer: H = 7.50 m above the launch point

Keep the unrounded value of v_iy (12.124…) in your calculator. Squaring it gives exactly 147.

Example 4: Range

A golf ball is hit at 30.0 m/s at 45.0° above level ground. How far away does it land?

A golf ball's dashed parabolic path from the ground back to the ground. A purple launch arrow is labeled 30.0 m/s at 45.0 degrees, and a double arrow along the ground is labeled delta x = ?.

Components: v_x = (30.0 m/s)(cos 45.0°) = 21.2 m/s and v_iy = (30.0 m/s)(sin 45.0°) = 21.2 m/s
Time to the top: t_up = (0 − 21.2 m/s) / (−9.8 m/s²) = 2.16 s
Total time: t_total = 2 × 2.16 s = 4.33 s
Range: Δx = v_x × t_total = (21.2 m/s)(4.33 s)
Answer: Δx = 91.8 m

Keep unrounded values in your calculator between steps; rounding early can change the last digit.

Big Ideas

Split v_i: v_x = v_i cos θ, v_iy = v_i sin θ
At the top: v_y = 0, but v_x remains
Hang time and height: vertical motion only
Range = v_x × total time

Back to the essential question: we split the launch velocity into components. The vertical part, v_i sin θ, is free fall: it sets the hang time and the maximum height, where v_y = 0. The horizontal part, v_i cos θ, stays constant, so range = v_x × hang time. Next up, we’ll study objects moving in circles with uniform circular motion.