Answer to Question #241973 in Physics for Kristine

Question

A stone is thrown upward with an initial velocity of 50 𝑓𝑡/𝑠. What will its maximum height be? When will it strike the ground? Where will it be in 1.25𝑠?

Answer 1

Given:

$v_0=50\:\rm ft/s$

$g=32.2\:\rm ft/s^2$

The maximum height reached by the stone

$$h_{\max}=\frac{v_0^2}{2g}=\frac{50^2}{2*32.2}=38.8\:\rm ft$$

The time of the motion of a stone

$$t=2*\frac{v_0}{g}=2*\frac{50}{32.2}=3.1\:\rm s$$

At the instant $t=1.25\:\rm s$ the position of a stone

$$h=v_0t-gt^2/2\\ =50*1.25-32.2*1.25^2/2=37.3\:\rm ft$$