Question

A ball is thrown vertically upward with a speed of 15 m/s.

A. what is its maximum height

B. when will it return

C. where will it be after 1 s ?

Answer 1

Given:

$v_0=\rm15\: m/s$

$g=9.8\:\rm m/s^2$

A. the maximum height

$$h_{\max}=\frac{v_0^2}{2g}=\frac{15^2}{2*9.8}=11.5\:\rm m$$

B. total time of motion

$$t=2*t_{up}=2*\frac{v_0}{g}=2*\frac{15}{9.8}=3.1\:\rm s$$

C. at the instant $t=1\:\rm s$

$$h=v_0t-gt^2/2\\ =15*1-9.8*1^2/2=10.1\:\rm m$$