Question

A ball is thrown vertically up with a certain velocity from the top of a tower of height 40 m. At 4.5 m above the top of the tower its speed is exactly half of that it will have at 4.5 m below the lop of the tower. Find the maximum height reached by the ball above the ground?

Answer 1

$v^2-\frac{v_1^2}{4}=2g\left(4.5\right)$ (For top)
$v_1^2-v^2=2g\left(4.5\right)$ (For bottom)
$v^2-\left(\frac{v^2+2g\left(4.5\right)}{4}\right)=2g\left(4.5\right)$
$\frac{3v^2}{4}=\left(\frac{5}{4}\right)\left(2g\right)\left(4.5\right)$
$v^2=\frac{450}{3}=150$
$v=\sqrt{150}m/s$
Maximum height from tower
$v^2=\left(2g\right)h^{\prime }$
$h^{\prime }=\frac{v^2}{2g}$
$=\frac{v^2}{20}$
$h^{\prime }=7.5$
Height of ball $=40+7.5=47.5$