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Q62. The horizontal range of a bullet fired with angle of projection 45° to the horizontal is 360 m. If it is fired from a lorry moving in the direction of bullet with the uniform velocity 18 km hr^–1 and with same elevation, what is the new range horizontal distance traveled by the bullet? (Take g = 10 ms^–2).

Dear Student

62. 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.
$$\begin{gathered} Range R = \frac{v^{2}Sin 2\theta }{g} \\ {v}^2=\frac{Rg}{Sin 2\theta }=\frac{360 x 10}{Sin 90}=3600 \\ v =60 m/s \\ Time of flight T =\frac{2v Sin\theta }{g}=\frac{2\times 60\times Sin 45}{10}=8.48s \\ Horizontal dis\tan ce covered by lorry x = v_{x}T \\ {v}_x= 18 km/h = 5m/s \\ x = v_{x}T= 5\times 8.48= 42.42 m \end{gathered}$$
So new range will be
360 m +42.42 m = 402.42 m

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