Question

A person standing on the edge of a cliff of height 40 m above the ground throws the first ball straight up with an initial velocity of 20 $\frac{m}{s}$, and then throws the second ball straight down with the same initial velocity 20 $\frac{m}{s}$. What are their respective velocities just before they hit the ground. (Take a = 10 $m/s^2$)

• A. $20\sqrt{2}$ $\frac{m}{s}$, $20\sqrt{2}$ $\frac{m}{s}$
• B. $20\sqrt{2}$ $\frac{m}{s}$, $20\sqrt{3}$ $\frac{m}{s}$
• C. $20\sqrt{3}$ $\frac{m}{s}$, $20\sqrt{2}$ $\frac{m}{s}$
• D. $20\sqrt{3}$ $\frac{m}{s}$, $20\sqrt{3}$ $\frac{m}{s}$

Answer 1

The correct option is D

$20\sqrt{3}$ $\frac{m}{s}$, $20\sqrt{3}$ $\frac{m}{s}$

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 = ${20}^2$ - 2 $\times$ 10 $\times$ s

s = 20 m

Total distance = 20 m + 40 m = 60 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 $\times$ 60

v = $\sqrt{1200}$ = $20\sqrt{3}$ m/s

For the second ball,

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

$v^2$ = ${20}^2$ + 2 $\times$ 10 $\times$ 40 = 1200

v = $20\sqrt{3}$ m/s