Question

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

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

Answer 1

The correct option is B

6.47 s

Separate the question into two parts:

Part 1: Ball going from roof to topmost height [final velocity = 0]

Using first equation of motion, v = u + at

0 = 19.6 – 9.8 $t_1$

$t_1$​ = 2 s

Using third equation of motion, $v^2=u^2+2as$

0 = ${19.6}^2$ – 2 $\times$ 9.8 $\times$ $s_1$

$s_1$ = 19.6 m

Part 2: From maximum height to ground [initial velocity = 0]

Total height = 78.4 + 19.6 = 98 m

Using second equation of motion, s = ut + $\frac{1}{2}a{t}^2$

98 = 0 + $\frac{1}{2}$ $\times$ 9.8 $\times$ $t_2^2$​

$t_2$​ = $\sqrt{20}$ s = 4.47s

Hence, total time taken = $t_1$​ + $t_2$​ = 2 + 4.47 = 6.47 s