Question

A ball is projected vertically upwards with a velocity of $25m{s}^{-1}$ from the bottom of a tower. A boy who is standing at the top of a tower sees it passes by him in the upward direction. He sees ball again at same height after $3$ sec when it is falling. Then the height of the tower is : $(g=10m{s}^{-2})$

• A. $5m$
• B. $10m$
• C. $15m$
• D. $20m$

Answer 1

The correct option is D $20m$

$s=ut-\frac{1}{2}g{t}^2$
$h=25\left(t\right)-\frac{1}{2}g(t{)}^2.....(1)$
$h=25(t+3)-\frac{1}{2}g(t+3{)}^2.....(2)$
$eq\left(2\right)-eq\left(1\right)\Rightarrow 0=25\left(3\right)-\frac{1}{2}g(t^2+9+6t-t^2)$
$25\left(3\right)=\frac{1}{2}g(9+6t)$
$75=5(9+6t)$ from (1)
$t=1sec$ hence from (1)

$s=25\left(1\right)-\frac{1}{2}g(1{)}^2$
$s=20m$