Question

Two bodies A and B are thrown simultaneously. A is projected vertically upwards with 20 ms speed
from the ground and B is projected vertically downwards from a height of 40 m with the same speed
and along the same line of motion. At what point do the 2 bodies collide? (s = 9.8m/s)​

Answer 1

Given:

Initial velocity of body A = 20 m/s

Height = 40 m

To find:

The height at which two bodies collide

Calculation:

Let the bodies collide at a height ‘h’ from the ground in time ‘t’ sec

Body A is thrown vertically upward, so the acceleration due to gravity on the body will be ‘-g’

From equations of motion

     $S=ut+\frac{1}{2} at^2$

     $h=ut-\frac{1}{2} gt^2$      ……(1)

Body A is thrown vertically downward, so the acceleration due to gravity on the body will be ‘g’

     $40-h=ut+\frac{1}{2} gt^2$   ……(2)

Add equations (1) & (2)

     $h+(40-h)=ut-\frac{1}{2} gt^2+(ut+\frac{1}{2} gt^2)$

     $40=2ut$      

     $40=2\times20\times t$

      $t=1sec$

Substitute the value of t in equation(1)

      $h=ut-\frac{1}{2} gt^2$

      $h=(20\times1)-\frac{1}{2} \times9.8\times1^2$

      $h=20-4.9$

      $h=15,1m$

Final answer:

The balls will collide at a height of 15 m from the ground in 1 sec.