Vertical throw upward

Beginner

Rising body decelerates at g, stops at the top, then falls.

Key relationH = u²/2g

A body thrown up decelerates at g, reaches zero velocity at the top, and then falls back. The motion is symmetric: it returns to the start with the same speed it left. Maximum height H = u²/2g, rise time u/g.

Worked example

Thrown up at 20 m/s: top at 20 m after about 2.04 s, back at 4.08 s.

Common mistake

At the top the velocity is zero but the acceleration is still g downward.

In more detail

When an object is thrown upward, it decelerates at g until its velocity is zero at the highest point, then falls back. The same equations apply, with the upward direction taken as positive.

How to solve it

Take upward as positive, so the initial velocity is u and a = −g.
At the highest point the velocity is zero, so use v² = u² − 2gh to find the height.
Time to the top is t = u/g.
The same equations describe the way down, with the signs reversed for displacement.

Worked examples

Example 1: A ball is thrown straight up at 15 m/s. How high does it rise, and how long does it take to reach the top?

Height: h = u²/(2g) = 225/19.6 = 11.5 m. Time: t = u/g = 15/9.8 = 1.53 s.

Example 2: For the same ball, when is it moving upward at 5 m/s?

Use v = u − gt, so 5 = 15 − 9.8t, giving t = 10/9.8 = 1.02 s after launch.

Checking your answer

By symmetry the ball returns to the launch point at speed 15 m/s, and the total time in the air is twice the time to the top, about 3.06 s.

Real-world uses

Estimating how high a ball rises when thrown straight up.
Calculating how fast a diver's hand should be moving to reach a set height.
A ball thrown straight up reaches a maximum height, and the same formula tells a physics student how high a catapult can launch a pumpkin.

Open the interactive lesson, with simulation and quiz →

Frequently asked questions

What is the velocity at the top of the motion?

Zero, for an instant, because the object momentarily stops before falling.

Is the speed on the way down the same as on the way up?

Yes, if air resistance is ignored. The object returns to the launch height with the same speed.