Question #94632

The muzzle speed for a Lee–Enfield rifle is 630 m s−1
. Suppose you fire this rifle at a target 700 m away and
at the same level as the rifle.
(a) In order to hit the target, you must aim at the barrel at a point above the target. How many
meters above the target must you aim? Pretend there is no air resistance.
(b) What will be the maximum height that the bullet reaches along its trajectory?
(c) How long does the bullet take to reach the target?

Expert's answer

(A)

We know that

u2sin(2θ)g=range=700putting all values6302sin(2θ)9.8=700sin(2θ)=0.0172839\frac{u^2\sin(2\theta)}{g}=range=700\\putting\ all\ values\\\frac{630^2\sin(2\theta)}{9.8}=700\\\sin(2\theta)=0.0172839


θ=0.5°tan(θ)=0.0087269=x700x=700×0.0087269=6.1 m\theta=0.5\degree\\\tan(\theta)=0.0087269=\frac{x}{700}\\x=700\times0.0087269=6.1\ m


So he should aim 6.1 m6.1\ m above the targe)


(B) max height = u2(sin(θ))22g=6302×(0.0087265)22×9.8=1.54207 m\frac{u^2(\sin(\theta))^2}{2g}=\frac{630^2\times(0.0087265)^2}{2\times9.8}=1.54207\ m


(C) Time of flight = 2usin(θ)g=2×630×0.00872659.8=1.122 sec\frac{2u\sin(\theta)}{g}=\frac{2\times630\times0.0087265}{9.8}=1.122\ sec


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