A bullet fires at an angle 30°with the horizontal hits the ground 3km away.
(a) what is the muzzle speed of the bullet?
(b) calculate the time taken by the bullet to reach the ground
(c) to cover maximum horizontal distance what will be the angle of projection?
(d) by adjusting angle of projection is it possible to hit a target 5km away?




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Angle of projection, α=30°
Range = 3 km = 3000m
let muzzle speed = u m/s

(a)Range =  \frac{ u^{2} sin(2 \alpha )}{g}

⇒3000 =  \frac{ u^{2} sin(60)}{9.8}

 u^{2} = \frac{3000*9.8}{sin(60)} = \frac{29400}{0.866} =33948.2
u= \sqrt{33948.2} =184.25 m/s

(b) Time of flight = t

t =  \frac{2usin( \alpha )}{g} = \frac{2*184.25*0.5}{9.8} =18.8s

(c)For maximum range, 

Range =  \frac{ u^{2} sin(2 \alpha )}{g} should be maximum.

or sin(2α) = 1
⇒sin(2α) = sin(90)
⇒ 2α = 90
⇒ α = 90/2 = 45°

(d) Maximum Range =  \frac{ u^{2} sin(2*45)}{g}  \frac{ u^{2} sin(90)}{g}  \frac{ u^{2}}{g}

⇒max. R = 3464m = 3.464 km <5 km

Since its maximum range is less than 5 km, by adjusting angle of projection, it is not possible to hit a target at 5 km.

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