Question

The balls are released from the top of a tower of height H at regular interval of time. When first ball reaches at the ground, the ${n^{th}}$ ball is to be just released and ${\dfrac{{(n + 1)}}{2}^{th}}$ ball is at some distance h from the top of the tower. Find the value of h.

Answer 1

Hint:In order to solve this question, we will use the general equations of motion and will establish the relation between height H and number of ball ${n^{th}}$ and then we will solve for the height h for the ${\dfrac{{(n + 1)}}{2}^{th}}$ ball.

Formula used:
Equation of motion is,
$S = ut + \dfrac{1}{2}a{t^2}$
where,
S is the distance covered by the body.
u is the initial velocity of the body.
a is the acceleration of the body and when body motion is under effect of gravity then $a = g$ called acceleration due to gravity.
t is the time taken by the body.

Complete step by step solution:
According to the question, we have given that balls are released at frequent interval from the height H. Let ‘t’ be the frequent interval of time at which balls are released so ${\dfrac{{(n + 1)}}{2}^{th}}$ ball will reach at a height of kH after the time $= (\dfrac{{n + 1}}{2} - 1)t = (\dfrac{{n - 1}}{2})t \\ $
Now, since balls are released under the effect of gravity so, $u = 0, a = g$.
Hence, using equation of motion $S = ut + \dfrac{1}{2}a{t^2}$ we get
$kH = \dfrac{1}{2}g{[(\dfrac{{n - 1}}{2})t]^2} \to (i)$

Also, we know for the height H we have,
$H = \dfrac{1}{2}g{[(n - 1)t]^2} \to (ii)$
Put the value H from equation (ii) to (i) we get,
\[k\dfrac{1}{2}g{[(n - 1)t]^2} = \dfrac{1}{2}g{[(\dfrac{{n - 1}}{2})t]^2} \\
\Rightarrow k\dfrac{1}{2}g{[(n - 1)t]^2} = \dfrac{1}{4}.\dfrac{1}{2}g{[(n - 1)t]^2} \\
\Rightarrow k = \dfrac{1}{4} \\ \]
So, ball covers a distance of $kH = \dfrac{H}{4}$
Hence the height h from the top of the tower is $= H - \dfrac{H}{2} = \dfrac{{3H}}{4} \\ $
Hence, the value of h is $\dfrac{{3H}}{4}$.

Note: It should be remembered that, while solving such problems make sure the sign conventions are properly used as if we take gravity as positive for a falling body then the distance from falling and velocity should also be taken as positive while in case of projecting a body vertically upwards, choose gravity as negative and other quantities as positive to avoid any mistake in calculation.