Question

A boy P stands on a top of a tower. Another boy Q throws a ball vertically up from the ground so that it just reaches P and the time taken is${t_o}$. The speed with which P should throw the ball vertically downwards so that it reaches the ground in time ${t_o}$ is:
A) $g{t_o}$
B) $2g{t_o}$
C) $\dfrac{3}{4}g{t_o}$
D) $\dfrac{{g{t_o}}}{4}$

Answer 1

The time taken in the ball is the same in both directions which means that the initial velocity when the ball is thrown from the point Q to P is more than the initial velocity of the ball when it is thrown from the point P to Q.

Formula used:
The formula of first relation of Newton’s law of motion is given by,
$v = u + at$
Where $v$ is the final velocity initial velocity is $u$ acceleration of the body is $a$ and time taken is $t$.

Complete step by step answer:
It is given in the problem that a boy P stands on a top of a tower another boy Q throws, a ball vertically up from the ground so that it just reaches P and the time taken is ${t_o}$ we need to find the speed with which P should throw the ball vertically downwards so that it reaches the ground in time ${t_o}$.
The formula of first relation of Newton’s law of motion is given by,
$v = u + at$
Now for motion from Q to P.
$\Rightarrow v = u + at$
Since the final velocity has to be zero and the time taken is zero therefore,
$\Rightarrow v = u + at$
On substituting the corresponding values,
$\Rightarrow 0 = u - g{t_o}$
$\Rightarrow u = g{t_o}$

The initial velocity for the ball travelling from the point Q to P is $u = g{t_o}$. The correct option for this problem is option (A).

Note:
The acceleration at the ball when the ball reaches from the point Q and P will be in the downwards direction and when the ball reaches from P to Q then the acceleration due to gravity is in the direction of motion in the downwards direction.