Question

A ball thrown by one player reaches the other in $2s$ the maximum height attained by the ball above the point of projection will be $(g = 10m/{s^2})$
A. $5m$
B. $13m$
C. $3m$
D. None of these

Answer 1

If we throw a ball upwards at a particular point it stops and it comes down. The velocity becomes zero at the maximum height that is a point where it turns around, If we throw a ball upwards it experiences a constant acceleration which is irrespective of the direction that is moving on it is known as acceleration due to since the Earth is not a perfect sphere the value depends on the place on the earth.

Complete step by step solution:
The ball will no longer go further after reaching the maximum height, at that instant, the motion of the ball changes the direction. Hence at this point, the magnitude of the velocity of the ball is zero.
From the projectile motion, the time is given by,
$T = \dfrac{{2u\sin \theta }}{{g\sec }} = 2\sec$
$g = 10 m/s^2$
$u\sin \theta = g$
Now squaring both sides,
${u^2}{\sin ^2}\theta = {g^2}$………..(1)
The ball’s maximum height is,
$H = \dfrac{{{u^2}{{\sin }^2 \theta}}}{{2g}}$……………(2)
Now substitute the value of ${u^2}{\sin ^2}\theta$ from equation (1) in (2) we get,
$H = \dfrac{{{g^2}}}{{2g}}$
After simplification it becomes,
$H = \dfrac{g}{2}$
Substituting the value of $g$,
$\Rightarrow H = \dfrac{{10}}{2} = 5m$
Therefore, the maximum height attained by the ball above the point of projection will be $5m$. So, option (A) is correct.

Note:

The acceleration would be the same as the acceleration because of gravity that is like if considering no air resistance acting downwards.
Consider the example: let upward direction be positive and downward as negative or vice versa as the final result has no effect of sign taken, when it reaches the maximum height the velocity becomes zero.