A ball is hit straight up from a height of 4 ft with an initial velocity of 65 ft/sec. Find the maximum height of the ball and the number of seconds to reach that height.
Thank you in advance for any help!
Initial velocity of ball $\displaystyle u = 65 \;\;ft/s$
acceleration due to gravity, $\displaystyle g = -9.81 \;\;m/s^2$(acting downwards)
at maximum height, the ball stops for a moment, and then returns back down.
so, final velocity, $\displaystyle v= 0$
suppose it takes time t to reach maximum height.
$\displaystyle v=u+gt$
$\displaystyle
0=65-9.81t$
$\displaystyle t= \frac{65}{9.81}=6.6$
it takes 6.6 s to reach max height.
Maximum height reached, h is given by
$\displaystyle h=ut + \frac{1}{2}gt^2$
$\displaystyle h=(65)(6.6) + \frac{1}{2}(-9.81)(6.6)^2$
$\displaystyle h=215.34\;\; ft$
total height from ground = 4 + 215.34 = 219.34 ft