The horizontal range of a bullet fired with angle of projection $45^{o}$ to the horizontal is $360 m$ . If it is fired from a lorry moving in the direction of bullet with the uniform velocity $18 k m$ $h^{- 1}$ and with same elevation, what is the new range horizontal distance traveled by the bullet (Take $g = 10 m s^{- 2}$ ).

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Solution

This is projectile motion, in which only the vertical component of initial velocity keeps changing, but the horizontal doesn't change at all.
Let the initial velocity be u.
Horizontal range for $45^{_{O}}$ $projection=\cfrac{u^2}g=360m$
$∴ u = \sqrt{\frac{360}{10}} = 6 m / s$
Time taken for entire flight is $T = \frac{2 u s i n θ}{g} = \frac{\sqrt{2} \times 6}{10}$ sec.
$18 k m / h r = \frac{18000}{3600} = 5 m / s$
Distance travelled by lorry $= 5 \times T = \frac{6}{\sqrt{2}} = 4.243$ metres
This distance will get added up to the old range.
New range = $364.243$ $m .$

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