Question

The horizontal range of a bullet fired with angle of projection 45° to the horizontal is 360 m . If it is fired from lorry moving in the direction of bullet with the uniform velocity 18 km/h and with same elevation , what is the new range horizontal distance traveled by the bullet ? ( Take g = 10 m/s² )

Ans -: 402.42 m​

Answer 1

Answer is in attachment

hope it helps

Answer 2

The new range will be the total of range and additional distance covered by lorry in time of flight T.
So we need to find out first time of flight and then additional distance covered by lorry in that time.
Range R = v2Sin 2θgv2=RgSin 2θ=360 x 10Sin 90=3600v =60 m/sTime of flight T =2v Sinθg=2×60×Sin 4510=8.48sHorizontal distance covered by lorry x = vxTvx= 18 km/h = 5m/sx = vxT= 5×8.48= 42.42 m
So new range will be
360 m +42.42 m = 402.42 m

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