Question

A stone is thrown upward with an initial speed of 48 ft/s.

A. what will its maximum height be ?

B. where will it be after 1 s ?

Answer 1

Given:

$v_0=48\:\rm ft/s$

$g=32.17\:\rm ft/s^2$

A. what will its maximum height be?

$$h_{\max}=\frac{v_0^2}{2g}=\frac{(48\:\rm ft/s)^2}{2*32.17\:\rm ft/s^2}=35.8\:\rm ft$$

B. where will it be after 1 s?

$$y=v_0t-gt^2/2=48*1-32.17*1^2/2=31.9\;\rm ft$$