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
proiected 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

Answer:

Explanation:vQ=vP=v=9.8m/S

tq=tp =t

Equation of motion for P

P=vt+gt^2/2

Equation of motion for particle at height Q

Q=-vt+gt^2/2_

P=2Q

From 1 ,2 and 3

2(-vt+gt^2/2) =vt +gt^2/2

-2vt+gt^2 =vt +gt^2/2

3v=gt/2

t=6v/g =6×9.8m/s÷9.8m/s^2

t=6sec