Question

5. The balls are released from the top of a tower of height H at
regular interval of time. When first ball reaches at the
(n+1)
ground, the n'h ball is to be just released and ball
is at same distance ‘h'from top of the tower. The value of h is.​

Answer 1

Answer:

h = H/4

Explanation:

The question is:

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

Solution:

Let the balls are release at time interval t

At time t second ball is released, at time 2t from the start, 3rd ball is released

Therefore at time $(\frac{n+1}{2}-1)t=\frac{n-1}{2}t$, the $\frac{n+1}{2}$th ball will be released

And, at time (n-1)t, the nth ball will be released

Time taken in reaching the first ball to the ground

using the second equation of motion

H = (1/2)gT²

$T=\sqrt{\frac{2H}{g}}$

T = (n - 1)t

Thus

$(n-1)t=\sqrt{\frac{2H}{g} }$

or, $t=\frac{1}{n-1} \sqrt{\frac{2H}{g}}$

We have to find the distance travelled by $\frac{n+1}{2}$th ball in time

(n-1)t - (n-1)t/2= (n-1)t/2

The distance travelled = h

$h=\frac{1}{2}g[\frac{(n-1)t}{2}]^2$

$\implies h=\frac{1}{2}g\frac{(n-1)^2t^2}{4}$

$\implies h=\frac{1}{2}g\frac{(n-1)^2}{4}\times (\frac{1}{n-1} \sqrt{\frac{2H}{g}})^2$

$\implies h=\frac{1}{2}g\frac{(n-1)^2}{4}\times \frac{1}{(n-1)^2}\times\frac{2H}{g}}$

$\implies h=\frac{H}{4}$

Therefore, the ball is at distance H/4 from the top of the tower.

Hope this helps.