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Question

A 24kg projectile is fired at an angle of 53 above the horizontal with an initial speed of 50m/s. At the highest point in its trajectory, the projectile explodes into two fragments of equal mass, the first of which fall vertically with zero initial speed. How far (in m) from the point of firing does the second fragment strike the ground? (Assume the ground is level)

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Solution

Here, m=24kg,u=50m/s
vx=dxdt=ucos53x=ucos53.t...(1) and
vy=dydt=gt+usin53y=12gt2+usin53.t...(2)
At highest point, vy=0, so gt+usin53=0t=50sin5310=4s and from (1) x=50cos50.4=120m
Momentum is conserved in exploration . If vx be the velocity of second particle , then mvx=(m/2)vxvx=2vx=2(50cos53)=60m/s
The total distance covered by the second fragment along x component is
x=60t+x=60(4)+120=360m

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