A particle p is projected from a point o with an initial velocity of 60m/s at an angle of 30° to the horizontal. at the same instant a second particle Q is projected in the opposite direction with an initial velocity of 50m/s from a point level with O and 100m from O.if the particle collide find the angle of projection of Q and find when the collision occurs.
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After time t when the particles collide they must be at the same height
h = vy t - 1/2 g t^2 final height of particle p
H = Vy t - 1/2 g t^2 final height of particle Q
vy = Vy since h = H
vy = 60 sin 30 = 30 m/s = Vy
Vy = 30 = 50 sin theta
theta = 36.9 deg the projection angle of Q
Vx = 50 * cos 36.9 = 40 m/s
vx = 60 * cos 30 = 52 m/s
(vx + Vx) * t = 100
t = 100 / 92 = 1.09 sec time for collision
Check: vy = 60 sin 30 = 30 m/s
Vy = 50 sin 36.9 = 30 m/s
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