Question

A ball is thrown upwards from the foot of a tower. The ball crosses the top of the tower twice after an interval of $4$ second and the ball reaches ground after $8$ seconds, then the height of tower in meters is $12\times x$. Then, find the value of $x$.

Answer 1

Suppose height of tower $=h$ after crossing the top $B$ of tower the ball reaches to a height point $P$ time taken by ball to go form $B$ to $P=\frac{4}{2}=2\text{sec}$.
Time taken by ball to go from $A$ to $P=\frac{2}{2}=4\text{sec}$.

Maximum height reached from foot of tower is$\frac{g{t}^2}{2}=\frac{10\times 4^2}{2}=80$
Maximum height reached from top of tower is$\frac{g{t}^2}{2}=\frac{10\times 2^2}{2}=20$
Height of the tower is $80-20=60=12x\Rightarrow x=5$