Question

A very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown upward when the previous one ball is at the maximum height. If the maximum height is $5m$ then the number of ball thrown per minute is ( $g = 10m{\sec ^{ - 2}}$ )
(A) $120$
(B) $80$
(C) $60$
(D) $40$

Answer 1

In order to solve this problem, we will use the general Newton’s equation of motions which are $v = u + at$ and ${v^2} - {u^2} = 2aS$ and will find the number of balls thrown in one second using these equations and then convert these number of balls in one minute.

Complete step-by-step solution:
Let us suppose a ball is thrown upward and its maximum height is given as $S = 5m$ let its initial velocity be $u$ and since the ball came to rest at its final point hence its final velocity $v = 0$ .
Putting these value in equation of motion ${v^2} - {u^2} = 2aS$
Here $a = - g$ due to gravity and it’s in downward direction
$0 - {u^2} = - 2 \times 10 \times 5$
${u^2} = 100$
$u = 10m{\sec ^{ - 1}}$
Now using equation of motion $v = u + at$ , we will find the time taken by ball to reach at height of $5m$ , here $a = - g$ due to gravity and it’s in downward direction:
$0 = 10 - 10t$
$t = 1\sec$ .
So, a ball takes $1\sec$ to attain a height of $5m$ and with this time every ball can be thrown upward in every second.
Hence, Number of balls thrown in one second is $1$
Number of balls thrown in $60\sec = 1\min$ is $60$
Hence, the correct option is (C) $60$.

Note: It should be remembered that, while ball or any other body is thrown upward against the force of gravity then acceleration due to gravity is taken as negative while thrown in the direction of force of gravity its magnitude is taken as positive and the other Newton’s equation of motion is $S = ut + \dfrac{1}{2}a{t^2}$ .