Question

A particle is projected vertically upwards from a point $O$ with initial speed $12.5\dfrac{m}{s}$. At the same instant another particle is released from rest at a point $10m$ vertically above $O$. Find the height above $O$ at which the particles meet?

Answer 1

Hint: In this question, we will use the concept of the second equation of motion. Firstly, we will put the values in the second equation of motion for the case of the particle travelling upwards and then again repeat this for the particle travelling downward. On comparing the two equations, we will reach at the answer.

Complete answer:
According to question, the diagram of this question will be:

The acceleration due to gravity $g = 9.8\dfrac{m}{{{s^2}}}$
Let the height at which the particles will meet be $l$
So, the distance travelled by the particle from the bottom $ = l$
So, the distance travelled by particle from the top $ = 10 - l$
Also, let the time in which they will meet be $t$
The second equation of motion is,
$s = ut + \dfrac{1}{2}a{t^2}$
For the particle travelling from the bottom,
$l = 12.5 \times t - \dfrac{1}{2} \times 9.8 \times {t^2}........(1)$
Similarly, the equation for the particle travelling from the top is,
$10 - l = 0 \times t + \dfrac{1}{2} \times 9.8 \times {t^2}........(2)$
On putting equation (1) in equation (2), we get,
$10 - \left( {12.5 \times t - \dfrac{1}{2} \times 9.8 \times {t^2}} \right) = 0 \times t + \dfrac{1}{2} \times 9.8 \times {t^2}$
$10 - 12.5 \times t + \dfrac{1}{2} \times 9.8 \times {t^2} = 0 \times t + \dfrac{1}{2} \times 9.8 \times {t^2}$
On further solving this equation,
$10 - 12.5t + 4.9{t^2} = 4.9{t^2}$
On cancelling $4.9{t^2}$ on both the sides,
$10 - 12.5t = 0$
$12.5t = 10$
On further solving, we get,
$t = \dfrac{{10}}{{12.5}}$
$t = 0.8s$
On putting the above value in equation (1), we get,
$l = 12.5 \times 0.8 - 4.9 \times {(0.8)^2}$
On solving the above equation, we get,
$l = 6.86m$
So, the height above $O$ at which the particles meet is $l = 6.86m$.

Note:
It is important to note that the value of acceleration due to gravity is the highest when the object is present on the surface of the earth. As we move at a height or as we move into the surface of the earth, the value of acceleration due to gravity decreases.