A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed, and neglect air resistance.

Asked by Abhisek | 1 year ago |  269

1 Answer

Solution :-

Bullet is fired at an angle = 30°

The bullet hits the ground at a distance of 3km = 3000 m

Horizontal range, R = \( \dfrac{ u^2 sin^2θ}{g}\)

3000 =\( \dfrac{u^2 sin 60°}{9.8}\)

u2 = \( \dfrac{ (3000 \times 9.8)}{\dfrac{\sqrt{3}}{2}}\)

\( \dfrac{ 2 (3000 \times 9.8)}{\sqrt{3}}=\dfrac{58800}{1.732}\) = 33949

Also, R’ = \( \dfrac{ u^2 sin2θ}{g}\) ⇒ 5000 

\( \dfrac{(33949 \times sin2θ)}{9.8}\)

Sin 2θ = \( \dfrac{ (5000 \times 9.8)}{33949 }\)

\( \dfrac{49000}{33949}\) =1.44

Since sine of an angle cannot be more than 1. Therefore, a target 5 km away cannot be hit

Answered by Pragya Singh | 1 year ago

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