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# The vertical height of the point P above the ground is twice that of Q...a particle is projected downwards with a speed of 5m/s from P and at the same time another particle is projected upwards with the same speed from Q....Both particles reach the ground simultaneously,then:a:PQ=b:time of flight of stones=

Jagjot singh
13 Points
2 years ago
Vq=Vo=V=5m/s (speed of particles)
tq=to=t (time taken to reach the ground)
P=vt+1/2at^2................(i)
Q=-vt+gt^2/2.................(ii)

P=2Q..............(iii)

Vt+1/2gt^2=2(-vt+gt^2/2)
Vt+gt^2/2=-2vt+gt^2

3vt=gt/2

t=8v/g
t=8×5/10
t=4sec
Karan Gupta
51 Points
2 years ago
Let height of point Q from ground = h
Then, height of point P from ground = 2h

For point Q, h  =  -5t + (1/2)gt^2.............................(1)
For point P, 2h =  5t + (1/2)gt^2..............................(2)

Subtracting (1) from (2), we get: h = 10t
Putting h = 10t in Equation (1),
10t= -5t + (1/2)*10*t^2
15 = 5t means time of flight T = 3 seconds
and Distance PQ = h = 10t = 30 m