Question

Three balls $ A, B, C $ are thrown from a height $ h $ with equal speeds upwards, downwards and horizontally respectively. What is the relation among speeds $ v_{A}, v_{B}, v_{c} $ with which they hit the ground?

• A. $ v_{A} = v_{B}= v_{C} $
• B. $ v_{A} >\, v_{C}>\, v_{B} $
• C. $ v_{A} = v_{B}>\, v_{C} $
• D. $ v_{A} <\, v_{C} <\, v_{B} $

Answer 1

Answer: (C)

Let the initial velocity of balls $A, B$ and $C$ are equal and its magnitude is $u$.

Since the ball A is projected with velocity $u$ in upward direction, so when it will come back to the projection point, its velocity remains same
So, the final velocity of ball A, when it hits the ground is given as
$v_{A}^{2}=u^{2}+2gh$
Hence, $v_{A}=\sqrt{u^{2}+2gh} \ldots\left(i\right)$
and the final velocity of ball $B$ is
$v_{B}=\sqrt{u^{2}+2gh}\dots\left(ii\right)$
But the initial vertical velocity of the ball $C$ is zero
So, $v^{2}_{c}=\sqrt{\left(0\right)^{2}+2gh}$
$\Rightarrow v_{c}=\sqrt{2gh}\dots\left(iii\right)$
Hence, it is clear from Eqs. $\left(i\right)$, $\left(ii\right)$ and $\left(iii\right)$, we get
$v_{A}=v_{B}>\,v_{c}$