Projectile motion equations with worked examples

A projectile problem is two one-dimensional problems that share a clock. Horizontally nothing pushes on the object, so its horizontal velocity never changes. Vertically, gravity pulls it down at g = 9.80 m/s^2 the whole time. Solve each direction on its own, and pass the time between them. Everything below ignores air resistance, as intro courses do, and follows OpenStax College Physics 2e section 3.4.

The equations

Split the launch velocity v0 at angle theta above the horizontal into components:

• v0x = v0 cos theta
• v0y = v0 sin theta

Then, with up positive and the launch point at x = 0:

• Horizontal: a_x = 0, v_x = v0x (constant), x = v0x t
• Vertical: a_y = -g, v_y = v0y - g t, y = y0 + v0y t - (1/2) g t^2

Three shortcuts follow when the projectile lands at the same height it was launched from (level ground):

• Time of flight: T = 2 v0 sin theta / g
• Range: R = v0^2 sin(2 theta) / g
• Maximum height: H = (v0 sin theta)^2 / (2g)

The level-ground shortcuts are the most misused formulas in the unit. If the launch and landing heights differ, they give the wrong answer, and you go back to the component equations.

Example 1: rolling off a table

A ball rolls off a 1.25 m high table at 2.00 m/s. How long is it in the air, and how far from the table does it land?

It leaves horizontally, so v0y = 0. The vertical motion is the same as dropping it:

• 1.25 = (1/2)(9.80) t^2, so t = sqrt(2 x 1.25 / 9.80) = 0.505 s
• Horizontal distance: x = 2.00 x 0.505 = 1.01 m

The horizontal speed has no effect on the time. A ball dropped from the same height at the same instant lands at the same moment. At impact, v_y = sqrt(2 x 9.80 x 1.25) = 4.95 m/s downward while v_x is still 2.00 m/s, so the impact speed is sqrt(2.00^2 + 4.95^2) = 5.34 m/s.

Example 2: an angled launch on level ground

A ball is kicked at 20.0 m/s at 30.0 degrees above level ground.

• v0x = 20.0 cos 30.0 = 17.3 m/s
• v0y = 20.0 sin 30.0 = 10.0 m/s
• Time of flight: T = 2 x 10.0 / 9.80 = 2.04 s
• Range: R = 20.0^2 x sin 60.0 / 9.80 = 35.3 m
• Maximum height: H = 10.0^2 / (2 x 9.80) = 5.10 m, reached at 10.0 / 9.80 = 1.02 s

Check a point along the way: at t = 1.00 s, v_y = 10.0 - 9.80 x 1.00 = 0.200 m/s. It is still rising, just barely, which fits a peak at 1.02 s.

Kick the same ball at 60.0 degrees and the range is the same 35.3 m, because sin 120 = sin 60. Complementary angles give equal ranges on level ground. The 60 degree kick stays up longer (3.53 s) and goes higher (15.3 m).

Example 3: why 45 degrees wins

On level ground, sin(2 theta) is largest when 2 theta = 90, so the range is greatest at 45 degrees. At 20.0 m/s that range is 20.0^2 / 9.80 = 40.8 m.

Range goes as v0^2, so doubling the launch speed quadruples the range: 40.0 m/s at 45 degrees gives 163 m. With air resistance, real ranges are shorter and the best angle drops below 45 degrees.

Example 4: a launch from a cliff

A stone is thrown at 15.0 m/s at 30.0 degrees above horizontal from the top of a 20.0 m cliff. Where does it land?

The landing point is 20.0 m below the launch point, so the level-ground shortcuts do not apply. Set up the vertical equation with the ground at y = 0 and the launch at y0 = 20.0 m:

• v0x = 15.0 cos 30.0 = 12.99 m/s, v0y = 15.0 sin 30.0 = 7.50 m/s
• Landing: 0 = 20.0 + 7.50 t - 4.90 t^2

That is a quadratic. The positive root is t = (7.50 + sqrt(7.50^2 + 2 x 9.80 x 20.0)) / 9.80 = (7.50 + 21.17) / 9.80 = 2.926 s, about 2.93 s. The other root is negative and belongs to a time before the throw, so it is thrown out.

• Range: x = 12.99 x 2.926 = 38.0 m
• Highest point: 7.50^2 / (2 x 9.80) = 2.87 m above the cliff top, so 22.9 m above the ground
• Impact: v_y = -21.17 m/s and v_x = 12.99 m/s, so the speed is 24.8 m/s at 58.5 degrees below the horizontal

Plugging these numbers into R = v0^2 sin(2 theta) / g gives 19.9 m, roughly half the real answer. That is the shortcut failing in exactly the case it was never meant for.

A quick check on the impact speed: energy conservation (unit 7 of the course) says v^2 = v0^2 + 2 g h = 225 + 392 = 617, and sqrt(617) = 24.8 m/s. The two methods agree.

Example 5: a drop from a moving plane

A package is released from a plane flying horizontally at 50.0 m/s, 490 m above the ground.

• Fall time: t = sqrt(2 x 490 / 9.80) = 10.0 s
• Horizontal distance: 50.0 x 10.0 = 500 m ahead of the release point

Without air resistance, the package stays directly below the plane the whole way down, because both keep the same 50.0 m/s horizontal velocity.

Mistakes that cost points

• sin theta instead of sin 2 theta. For 30.0 m/s at 45.0 degrees, the right range is 91.8 m. Using sin 45 gives 64.9 m.
• Using the level-ground formulas from a height. Example 4 shows how far off that goes.
• Saying the speed at the top is zero. Only v_y is zero at the peak. The object still moves at v0x, and the acceleration is still 9.80 m/s^2 downward.
• Plugging v0 into a one-direction equation. Each direction gets its own component. The full launch speed never goes directly into x = v0x t or v_y = v0y - g t.
• Losing a sign. Pick up as positive, write a_y = -9.80 m/s^2, and keep it that way through the whole problem.

Practice

The projectile motion calculator runs any launch speed, angle and height and shows each step, so you can check your hand work. Projectile motion leans on one-dimensional kinematics, which is covered in the four kinematic equations and when to use each, and the kinematics calculator solves those. For where this sits in the course, see what's in College Physics I, and for exam prep, how to study for a physics exam. The list-your-knowns habit carries into chemistry too, where the ideal gas law is one equation with four quantities to solve for.

Projectile motion is unit 3 of Encodr's free College Physics I deck, with the equation sheet at the start of every chapter.