Answer to Question #330168 in Calculus for Gaile

Question

A stone is thrown vertically upward from the ground with an initial velocity of 40 ft/sec. 

 What is the height of the ball when its velocity is one-half the initial velocity?

Answer 1

$v\left( t \right) =v_0-gt\\x\left( t \right) =v_0t-\frac{gt^2}{2}\\v\left( t \right) =\frac{1}{2}v_0\Rightarrow v_0-gt=\frac{1}{2}v_0\Rightarrow t=\frac{v_0}{2g}\\x\left( t \right) =v_0\cdot \frac{v_0}{2g}-\frac{{gv_0}^2}{8g^2}=\frac{3{v_0}^2}{8g}=\frac{3\cdot 40^2}{8\cdot 32.174}=18.6486ft$