Question

The vertical height of P above the ground is twice that of Q. A particle is projected downward with a speed of $9.8\,m/s$ from P and simultaneously another particle is projected upward with the same speed of $9.8\,m/s$ from Q. Both particles reach the ground simultaneously. The time taken to reach the ground is
A. $3\sec $
B. $4\sec $
C. $5\sec $
D. $6\sec $

Answer 1

Hint: Here a particle is projected from a certain height and reaches the ground in a certain time and we have to find that time for that we need to use the third equation of motion which will provide the relation between the height of the particle and the time it reaches the ground.

Formula used:
The expression of third equation of motion,
$s = ut + \dfrac{1}{2}a{t^2}$
Where, $s$ is the displacement, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

Complete step by step solution:
Let the time taken to reach the ground be t.
First start with point P,
Let the height at point P be 2h.
So distance , $s = - 2h$
Speed of the particle when it is projected downward is,
$u = 9.8\,m/s$...........(as given in the question)
Now using the third equation of motion:
$s = ut + \dfrac{1}{2}a{t^2}$..........($a = 9.8\,m/{s^2}$ acceleration due to gravity)
Putting all the values, we get;
$ - 2h = 9.8t + \dfrac{1}{2}(9.8){t^2}$..........(equation 1)

Now for point Q, height will be half of P that is h.
So distance, $s = - h$
Speed of the particle when it is projected downward is the same as P that is $9.8\,m/s$.
Putting all the value in third equation of motion, we get;
$ - h = 9.8t + \dfrac{1}{2}(9.8){t^2}$.........(equation 2)
Multiply equation 2 by 2,
$ - 2h = 2 \times 9.8t + 2 \times \dfrac{1}{2}{t^2}$
By solving,
$ - 2h = 19.6t + {t^2}$.........(equation 3)
Subtracting equation 1 from 3, we get;
$9.8t + \dfrac{1}{2}{t^2} = 0$
Solving we get,
$\therefore t = 4$

Hence the correct answer is option B.

Note: Here the speed with which the object is projected is same from both the point P and Q but is not the same in all the cases if the speed given will be different then the answer will also get changed so be careful about this. Also check the height of the particle from which it is projected from both the points.