Question

A ball is thrown vertically upwards from the top of a building of height $H.$ It goes to a height of $20m$ and then comes back to the ground. Time taken by the ball to reach the maximum height from the top of the building is half of the time taken to reach the ground from the maximum height. What is the height of the building? (Take $g=10m{s}^{-2}$)

• A. $40m$
• B. $60m$
• C. $80m$
• D. $100m$

Answer 1

The correct option is B $60m$

$T_{OA}=\frac{1}{2}{T}_{AB}....\left(i\right)$

$OA$

$v^2-u^2=2gh$

$0-u^2=2\times -10\times 20$

$u^2=400$

$u=20m{s}^{-1}$

$v=u+gt$

$0=20-10\times t$

$t_{AO}=2\sec$

$AB$

$v=u+gt$

$v=10t$

$t_{AB}=\frac{v}{10}...\left(ii\right)$

$2=\frac{1}{2}+\frac{v}{10}\Rightarrow v=40m{s}^{-1}$

$v^2-u^2=2gh$

$\frac{{40}^2}{2\times 10}=H_{AB},H_{AB}=80m$

Building height $=80-20$

$=60m.$