Question

a ball player catches a ball 3.53 seconds after throwing it vertically upward.
- with what speed did he throw it?
-what height did it reach?

Answer 1

Z coordinate of the ball depends on time as

$$z(t) = v_0 t -\frac{gt^2}{2}$$

where g is $9.8 \,\text{ms}^{-2}$ - free fall acceleration., $v_0$ is initial speed of the ball.

a ball player catches a ball $\tau = 3.53$ seconds after throwing it, therefore

$$z(\tau) = v_0 \tau -\frac{g\tau^2}{2}=0 \Rightarrow v_0= \frac{g\tau}{2} = 17.3\,\text{ms}^{-1}$$

The Ball reached maximum heigth at mid time between throwing ball and catching it:

$$H = z(\tau/2) = \frac{g \tau^2}{8}=15.3\,\text{m}$$