A particle is projected from the ground with an initial speed of v ⁡ at an angle θ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is

Question

A particle is projected from the ground with an initial speed of v ⁡ at an angle θ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is, v ⁡ 2 1 + 2 cos 2 θ, v ⁡ 2 1 + cos 2 θ, v ⁡ 2 1 + 3 cos 2 θ, v ⁡ cos θ

MARKS App · Accepted Answer

Average velocity = displacement time v ⁡ av = H 2 + R 2 4 T / 2 ...(i) Here, H = maximum height = v ⁡ 2 sin 2 θ 2 g R = range = v ⁡ 2 sin 2 θ g and T = time of flight = 2 v ⁡ sin θ g Substituting in Eq. (1) we get v ⁡ av = v ⁡ 2 1 + 3 cos 2 θ

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