Question

A ball is thrown vertically upwards with a velocity of $20\, ms ^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is $25\, m$ from the ground. The height to which the ball rises from the ground and the time taken before the ball hits the ground are (Take $g = 10\, ms ^{-2}$ )

• A. 20 m , 2 s
• B. 40 m , 3 s
• C. 40 m , 5 s
• D. 45 m , 5 s

Answer 1

Answer: (D)

Taking point of launch as origin and up as + ve, using $v^{2}=u^{2}+2 a s$
$0^{2}=\left(20\, ms ^{-1}\right)^{2}+2\left(-10\, ms ^{-2}\right) \cdot H_{\max }$
$\Rightarrow H_{\max }=\frac{20^{2}}{20}=20\, m$
Hence, from the ground the maximum height reached $=25 m +20 m =45 m$
Further, for time to reach ground
$s=-25 m , u=+20 ms ^{-1}, a=-10 ms ^{-2}$
Using, $s=u t+\frac{1}{2} a t^{2} \Rightarrow -25=20 t-5 t^{2}$
$5 t^{2}-20 t-25=0$ or $5 t^{2}-25 t+5 t-25=0$
$5 t(t-5)+5(t-5)=0 \Rightarrow (5 t+5) \cdot(t-5)=0$
$t=-1 s$ or $t=5 s$
Rejecting -ve value of time, we get $t=5 s$.