Using the usual connotation, we can say that: y = usin(θ)t-gt^2/2 = vsin(Φ)t-gt^2/2 or usin(θ)=vsin(Φ). Use the given values to obtain: 60*1/2 =50sin(Φ). In other words, sin(Φ)= 3/5.
The second projectile must therefore be launched at angle Φ~=36.87° to the horizontal.
Next,assume that the two projectiles meet after t secs. Then:t(ucos(θ))+t(vcos(Φ))= 100 or t(60√3/2 + 50*4/5)=100 which yields t~ =1.0874 sec
A particle P is launched from a point O with an initial velocity of 60m/s at an angle of 30 degrees to the horizontal. At the same time, a second particle Q is projected in the opposite direction with initial speed of 50m/s from a point level with point O and 100m from O. You are requested to determine the angle of projection of particle Q needed if the particles collide. Also determine when the collision occurs.
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