Question

A ball was thrown with a velocity of 55 ft/s upwards. If caught at the same level as it was thrown, how high did the ball rise? And how long was the ball in the air?

Answer 1

Let's first find the time that the ball takes to reach the maximum height:

$$v=v_0+gt,$$ $$0=v_0+gt,$$ $$t=-\dfrac{v_0}{g}=-\dfrac{55\ \dfrac{ft}{s}}{-32.17\ \dfrac{ft}{s^2}}=1.71\ s.$$

Then, we can find the maximum height reached by the ball from the kinematic equation:

$$y_{max}=v_0t+\dfrac{1}{2}gt^2,$$ $$y_{max}=55\ \dfrac{ft}{s}\times1.71\ s+\dfrac{1}{2}\times(-32.17\ \dfrac{ft}{s^2})\times(1.71\ s)^2,$$ $$y_{max}=47\ ft.$$

Finally, we can find the total flight time of the ball in the air as follows:

$$t_{flight}=2t=2\times1.71\ s=3.42\ s.$$