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Q.

A projectile is projected with a speed 20 m/s from the floor of a 5 m high room as shown. The maximum horizontal range of the projectile is found to be 20n m. Find n [Take g = 10 ms-2].

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Detailed Solution

for Rmax,θ=45∘⇒Hmax=(20)2sin2⁡45∘2×10=10m∵Hmax allowed =5m,⇒5=(20)2sin2⁡θ2×10⇒θ=30°⇒Rmax=(20)2sin⁡60∘10m=203m.
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