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A person on a moving truck with velocity 14.7 m/s on horizontal road throws a ball in such way that it return to the truck after moving 58.8m find the speed and angle of projection as seen from the road A person on a moving truck with velocity 14.7 m/s on horizontal road throws a ball in such way that it return to the truck after moving 58.8m find the speed and angle of projection as seen from the road
Velocity=14.7 m/s Time taken by the truck to cover 58.8=t=S/v=58.8/14.7 =4secSince ball returns back to truck its angle of projection is vertically upwards with respect to truck.u=speed of projection.t=4sh=0h=4u-1/2 x 9.8x160=4u-78.44u=78.4u=78.4/4=19.6 m/sb) As seen from the road the speed of ball will be resultant of 2 speeds verticalspeed =u=19.6 m/sHorizontal speed given by truck=14.7m/s⇒√u²+v²=√19.6²+14.7²=24.5 m/sThus angle seen from road:Tan⁻¹[19.6/14.7] =Tan⁻¹1.33=Tan⁻¹[4/3]=53°with horizontal.
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