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:

$v_0=50\:\rm ft/s=15\:\rm m/s$

$t_1=1.25\: \rm s$

The maximum height

$$h_{\max}=v_0^2/2g=15^2/2*9.8=12\:\rm m$$

At the instant 1.25 s the vertical position:

$$y=v_0t-gt^2/2\\ =15*1.25-9.8*1.25^2/2=11 \:\rm m$$