Answer to Question #139797 in Physics for Kay

Question

A baseball is thrown upward at 20 m/s. It takes 15.5 s before the baseball strikes the ground. Neglect air
resistance.

a. How high above the ground was the baseball when it was thrown?
b. What is the velocity of the baseball just before it hits the ground?

Answer 1

(a) The time of upright movement of a baseball

$$t_1=\frac{v_0}{g}=\rm\frac{20\: m/s}{9.8\: m/s^2}=2.04\: s$$

The time of downward movement of a baseball

$$t_2=t-t_1=\rm 15.5-2.04=13.5\: s$$

The upright distance traveled by a baseball

$$h_{1}=\frac{v_1^2}{2g}=\rm\frac{(20\: m/s)^2}{2\times 9.8\: m/s^2}=20.4\: m$$

The downward distance

$$h_2=\frac{gt^2}{2}=\rm\frac{9.8\: m/s^2\times (13.5\: s)^2}{2}=887.6\: m$$

Hence, the initial height of a baseball above the ground

$$H=h_2-h_1=\rm 887.6\: m-20.4\: m=867\: m$$

(b) The velocity of the baseball just before it hits the ground

$$v_2=gt_2=\rm 9.8\: m/s^2\times 13.5\; s=132\; m/s$$