The range of riffle bullet is 1000m when θ is the angle of projection.if the bullet is fired with the same angle from a car travelling at 36km/h towards the target show that the range will be increased by ((1000)/7).(√(tanθ)) m.


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Question Number 50915 by peter frank last updated on 22/Dec/18

$T h e r a n g e o f r i f f l e b u l l e t$ $i s 1000 m w h e n θ i s t h e a n g l e$ $o f p r o j e c t i o n . i f t h e b u l l e t$ $i s f i r e d w i t h t h e s a m e$ $a n g l e f r o m a c a r t r a v e l l i n g$ $a t 36 k m / h t o w a r d s t h e t a r g e t$ $s h o w t h a t t h e r a n g e w i l l$ $b e i n c r e a s e d b y$ $\frac{1000}{7} . \sqrt{t a n θ} m .$
	Answered by mr W last updated on 22/Dec/18

	$c a s e 1 : f i r e d f r o m t h e g r o u n d$ $u \sin θ t - \frac{1}{2} {g t}^{2} = 0 \Rightarrow t = \frac{2 u \sin θ}{g}$ $L_{1} = u \cos θ t = \frac{2 u^{2} \sin θ \cos θ}{g} = \frac{u^{2} \sin 2 θ}{g}$ $\Rightarrow u = \sqrt{\frac{{g L}_{1}}{\sin 2 θ}}$ $c a s e 2 : f i r e d f r o m t h e c a r w i t h s p e e d v$ $u \sin θ t - \frac{1}{2} {g t}^{2} = 0 \Rightarrow t = \frac{2 u \sin θ}{g}$ $L_{2} = (u \cos θ + v) t = u \cos θ t + v t = L_{1} + v t = L_{1} + \frac{2 u v \sin θ}{g}$ $Δ L = L_{2} - L_{1} = \frac{2 u v \sin θ}{g}$ $Δ L = \frac{2 v \sin θ}{g} \sqrt{\frac{{g L}_{1}}{\sin 2 θ}}$ $Δ L = v \sqrt{\frac{4 \sin^{2} θ {g L}_{1}}{g^{2} 2 \sin θ \cos θ}}$ $\Rightarrow Δ L = v \sqrt{\frac{2 \tan θ L_{1}}{g}}$ $w i t h v = 36 k m / h = 10 m / s, g = 9.81 m / s^{2}, L_{1} = 1000 m$ $\Rightarrow Δ L = 10 \sqrt{\frac{2 \tan θ 1000}{9.81}}$ $\Rightarrow Δ L = \frac{1000}{\sqrt{49.05}} \sqrt{\tan θ} \approx \frac{1000}{7} \sqrt{\tan θ}$
	Commented by peter frank last updated on 22/Dec/18

	$t h a n k y o u$
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