The-range-of-riffle-bullet-is-1000m-when-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-incre

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

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