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+1 vote
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in Computer by (20 points)
edited by

Derive an expression for range of a projectile and for what angle of projection horizontal range becomes maximum?

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1 Answer

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Maximum horizontal range :-

sin 2theta = 1

sin 2theta = sin 90°

theta = 45°

R = u2 sin2 (45°)/g

R = u2 sin 90° /g

R = u2/g

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