Question

Two balls are projected simultaneously with the same speed $u$ from the top of a tower, one vertically upwards and the other vertically downwards. Their respective times of the journeys are $t_1$ and $t_2$. At the time of reaching the ground, the ratio of their final velocities is:

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

Answer 1

Answer: (A)

$1:1$

Using: $v^2=u^2+2aS$

where, $v$ final velocity, $u$ initial velocity $a$ is acceleration and $S$ is displacement of the particle.

For particle 1: $v_1^2=u_1^2+2(-g)(-h)$ taking upward direction positive

For particle 2: $v_2^2=u_2^2+2(-g)(-h)$

as $u_1=u_2$ and displacement of both the particle is equal to height of tower, their velocities at time of reaching ground will be equal.