Question

Raj standing on the edge of a cliff of height 35 m above the ground throws the first ball straight up with an initial velocity of 30 $\frac{m}{s}$, and then throws the second ball straight down with the same initial velocity 30 $\frac{m}{s}$. What are their respective velocities just before they hit the ground. (Take a = 10 $m/s^2$)

• A. $9\sqrt{2}$ $\frac{m}{s}$, $9\sqrt{2}$ $\frac{m}{s}$
• B. $16\sqrt{2}$ $\frac{m}{s}$, $16\sqrt{2}$ $\frac{m}{s}$
• C. 18 $\frac{m}{s}$, 18 $\frac{m}{s}$
• D. 40 $\frac{m}{s}$, 40 $\frac{m}{s}$

Answer 1

The correct option is D 40 $\frac{m}{s}$, 40 $\frac{m}{s}$

For the first ball, consider when it is going upwards to reach the maximum height.

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

0 = ${30}^2$ - 2 $\times$ 10 $\times$ s

s = 45 m

Total distance = 45 m + 35 m = 80 m

Now, consider second part where ball from topmost point starts falling down and reaches the ground

$v^2=u^2+2as$

$v^2$ = 0 + 2 $\times$ 10 $\times$ 80

v = $\sqrt{1600}$ = 40 m/s

For the second ball,

$v^2=u^2+2as$

$v^2$ = ${30}^2$ + 2 $\times$ 10 $\times$ 35 = 1600

v = 40 m/s