Question

A boy is standing on the edge of the roof at height of $78.4m$. He throws a ball vertically upwards with velocity of $19.6m{s}^{-1}$. After how much time the ball will reach the ground? (Take $g=9.8m{s}^{-2}$)

• A. $5.42s$
• B. $6.47s$
• C. $4.47s$
• D. $3.42s$

Answer 1

The correct option is B

$6.47s$

Given, height of the roof, $h=78.4m$
Consider two cases:
Case 1: Ball going from roof to topmost height
Given,
initial velocity, $u_1=19.6m{s}^{-1}$, final velocity, $v_1=0$ and acceleration, $a=-9.8m{s}^{-2}$ (negative sign because the ball is moving up)
Let the time taken be $t_1$ and distance covered be $s_1$.
Using first equation of motion, $v_1=u_1+a{t}_1$.
We have, $0=19.6–9.8t_1$
$\Rightarrow$ $t_1=2s$

Using third equation of motion, $v_1^2=u_1^2+2a{s}_1$
$0={19.6}^2–2\times 9.8\times s_1$
$\Rightarrow$ $s_1=19.6m$

Case 2: From maximum height to ground
In this case, initial velocity $u_2=0$
Total height, $s_2=s_1+h=78.4+19.6=98m$
Using second equation of motion, $s_2=u_2{t}_2+\frac{1}{2}a{t}_2^2$
$\Rightarrow 98=0+\frac{1}{2}\times 9.8\times t_2^2$​
$\Rightarrow t_2$​ = $\sqrt{20}=4.47s$
Hence, total time taken $t=t_1+t_2=2+4.47=6.47s$