Question

A boy throws n balls per second at regular time intervals. When the first ball reaches the maximum height he throws the second one vertically up. The maximum height reached by each ball is
(A) $\dfrac{g}{{2{{(n - 1)}^2}}}$
(B) $\dfrac{g}{{2{n^2}}}$
(C) $\dfrac{g}{{{n^2}}}$
(D) $\dfrac{g}{n}$

Answer 1

Hint: In order to solve this question, we will first find the initial velocity of the ball using newton’s equation of motion and then we will solve for the maximum height of the ball using newton’s equation of motion.

Formula used:
If u and v are the initial and final velocity of a body and a and t be the acceleration and time taken by the body then Newton’s equation is given as
$v = u + at$
${v^2} - {u^2} = 2aS$
where S is the distance covered by the body.

Complete answer:
We have given that A boy throws n balls per second at regular time intervals which means the time difference between two successive throws is $t = \dfrac{1}{n}$ which means a ball takes time $t = \dfrac{1}{n}$ seconds to reach the highest point vertically.

Now, after reaching highest point its final velocity will be $v = 0$ and acceleration will be $a = - g$ then its initial velocity is calculated using formula $v = u + at$ we get,
$0 = u - g\left(\dfrac{1}{n}\right) $
$\Rightarrow u = \dfrac{g}{n}$

Now, Using formula ${v^2} - {u^2} = 2aS$ and solving for maximum height S, we have
${0^2} - \dfrac{{{g^2}}}{{{n^2}}} = 2( - g)S $
$S = \dfrac{g}{{2{n^2}}} $
So, maximum height attained by the ball is $S = \dfrac{g}{{2{n^2}}}$

Hence, the correct option is (B) $\dfrac{g}{{2{n^2}}}$

Note: It should be remembered that, whenever a body is thrown vertically upward or it falls freely, acceleration due to gravity acts on it and for upward motion, this acceleration is taken as negative while for downward motion acceleration is taken as positive.