Question

A particle is dropped from the top of a tower $h$ meter high and simultaneously a particle is projected upwards from bottom. They meet when the upper one has descended $\dfrac{1}{n}$ of the distance. Find the initial velocity of the lower particle ?

Answer 1

Hint: In this question, we will apply the second speed equation $s = ut + \dfrac{1}{2}g{t^2}$ where s is the distance covered, t is the time taken and u is the initial velocity of the particle for both the particles. For the particle dropped from the top, the initial velocity will be taken as zero while the distance to be $\dfrac{h}{n}$ where h is the total height. For the other particle, the distance travelled will be taken as $s = h - \dfrac{h}{n}$ . Now since the two bodies meet at a point in their journeys, the time in air is the same for both the particles. Hence, we will substitute the value of time calculated from the journey of one particle in the other’s equation to get the final answer.

Complete step by step answer:
Let the total height be $h$. For the particle dropped from the top, the initial velocity is given as $u = 0\,m\,{s^{ - 1}}$. The distance travelled is given to be $\dfrac{1}{n}$ of the total height.Hence, we can say that,
$s = \dfrac{h}{n}$
where $s$ is the distance covered.
Using the second speed equation, we have,
$s = ut + \dfrac{1}{2}g{t^2}$
where $s$ is the distance covered, $t$ is the time taken and $u$ is the initial velocity of the particle.
Substituting the known values, we have,
$\dfrac{h}{n} = 0 + \dfrac{1}{2}g{t^2}$
$ \Rightarrow \dfrac{h}{n} = \dfrac{1}{2}g{t^2}$
This can be rewritten as
$t = \sqrt {\dfrac{{2h}}{{ng}}} \,\,\,\,\,\,\,\,\,\,\,\,\,.......(1)$

For the particle thrown upwards, we have the initial velocity be $u\,m\,{s^{ - 1}}$. The distance travelled by the body dropped from the top is given to be $\dfrac{1}{n}$ of the total height. Hence, we can say that $s = \dfrac{h}{n}$ where s is the distance covered. So, the distance travelled by the body thrown from the ground is given as $s = h - \dfrac{h}{n}$
This can be rewritten as $s = h(\dfrac{{n - 1}}{n})$
Using the second speed equation, we have,
$s = ut + \dfrac{1}{2}g{t^2}$
where s is the distance covered, t is the time taken and u is the initial velocity of the particle.
Substituting the known values, we have,
$h(\dfrac{{n - 1}}{n}) = ut - \dfrac{1}{2}g{t^2}$

Now since the two bodies meet at a point in their journeys, the time in air is the same.So, the time in equation (1) must satisfy the above equation as well.Hence, substituting in the equation, we get,
$h(\dfrac{{n - 1}}{n}) = u\sqrt {\dfrac{{2h}}{{ng}}} + \dfrac{1}{2}g \times \dfrac{{2h}}{{ng}}$
Cancelling the common terms, we get,
$h(\dfrac{{n - 1}}{n}) = u\sqrt {\dfrac{{2h}}{{ng}}} - \dfrac{h}{n}$
Rearranging the terms we get,
$h = u\sqrt {\dfrac{{2h}}{{ng}}} $
This can be rewritten as,
$u = \dfrac{h}{{\sqrt {\dfrac{{2h}}{{ng}}} }}$
$\therefore u = \sqrt {\dfrac{{ngh}}{2}} $

Therefore, the initial velocity of the lower particle $\sqrt {\dfrac{{ngh}}{2}}$.

Note:For the particle dropped from the top of the tower, we take acceleration to be positive since the direction of acceleration is the same as the direction of motion which is towards the ground. However, for the particle thrown upwards, the direction of acceleration acting on the particle is just the opposite to the direction of motion. Hence, we took a negative sign along with the acceleration in the speed equation. So, in order to get correct answers, we must take care of the signs we apply.