Question

During practice, a basketball was tossed upward with a speed of 15.0 m/s. It is caught at the same distance above the ground. (a) How high did the ball rise? (b) How long did the ball remain in the air?

Answer 1

Let us choose the origin, which coincides with the initial position of the ball. Then the equations of vertical motion are $y(t) = v_0 t - \frac{g t^2}{2}$, $v(t) = v_0 - g t$.

At highest point, the speed is zero, hence $v(t') = v_0 - g t'$, from where the time to reach that point is $t' = \frac{v_0}{g}$. Hence, the ball rose at height $y(t') = \frac{v_0^2}{g} - \frac{g}{2} \frac{v_0^2}{g^2} = \frac{v_0^2}{2 g} \approx 11.47 m$.

To find the time the ball remains in the air, equate its y coordinate to zero, and solve for time:

$y(t) = v_0 t - \frac{g t^2}{2} = t(v_0 - \frac{g t}{2}) = 0$, from where $t = 0$ (initially the ball is at the origin) or $t = \frac{2 v_0}{g} \approx 3.1 s$. Hence, the ball remains $3.1 s$ in the air.