Question

A man standing on the edge of a cliff throws a stone straight up with initial speed $u$ and then throws another stone straight down with same initial speed $u$ from the same position. Find the ratio of speeds, the stones would have attained when they hit the ground at the base of the cliff?

• A. $2:1$
• B. $1:2$
• C. $1:1$
• D. $3:1$

Answer 1

Answer: (C)

$1:1$

Speed of a particle (magnitude of velocity) is the same at same height.

So, when the man throws the ball with speed $u$, it will return to him with the same speed $u$.

$v=u+at$

$\Rightarrow 0=u-gt$ (final velocity on reaching the top is $0$,negative sign due to downward acceleration)

$\Rightarrow u=gt$ ---------$\left(i\right)$

While coming down $v=u+at$

$\Rightarrow -v=0-gt$ (negative sign due to downward direction)

$\Rightarrow v=gt$ ------$\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$,

$\left|v\right|=\left|u\right|$

So when man throws the stone upward with speed $u$ and downward with speed $u$, the initial velocity (from the cliff) will be the same. Hence, this will result in equal final velocity.

Hence the ratio is $1:1$