A Ball Thrown Upward

Example 2-12 A Ball Thrown Upward

A softball player throws a ball straight upward with a velocity of 17 m/s, and catches it exactly where it left her hand.

a.
How long does the ball remain in the air?

b.
How high does it go?

Solution

Assumptions. We take the origin to be the point where the ball leaves her hand. We assume that effects like air resistance are negligible, so that the ball is in free fall.

Restating the problem. The ball remains in the air until it is back where it started—at y = 0. Part a therefore asks

Part b, similarly to Example 2-9, asks

What we know/what we don't. In addition to this restatement of the question of the question, we know that

and because the ball is falling freely,

Choice of approach. Again, we may apply the constant acceleration equations of motion. For part a, the equation relating y and t is Equation 2-11, written as . If y is known, t will be the only unknown.
For part b, we apply Equation 2-12, written as for vertical motion, to obtain a one-step solution.

The mathematical solution.

a.
When y = y_o = 0 and a = −g, Equation 2-11 becomes

This is a quadratic equation that can be solved for t by factoring:

so that either one or the other factor must be 0:

in which case, solving for t gives

There are two solutions for t, and both have meaning in the actual situation. The ball is at y = 0 first at t = 0 when it leaves the player's hand, then again at t = 3.5 s when it returns to her glove. The second solution answers the question of how long the ball remains in the air. (In fact, for any value of y, Equation 2-11 yields two solutions for t because as Figure 2-21a shows, the ball passes each value of y, other than the maximum value, both on the way up and on the way down.)

b.
With y_o = 0 and a = −g, becomes . Proceeding much as in Example 2-9, when v = 0 we get

Related homework: Problems 2-49 and 2-102.