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Q.

A body is projected with velocity u at an angle of projection θ with the horizontal. The direction of velocity of the body makes angle 30o with the horizontal at t = 2 s and then after 1 s it reaches the maximum height. Then

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a

u=203ms−1

b

θ=60∘

c

θ=30∘

d

u=103ms−1

answer is A.

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Detailed Solution

Time of ascent =2+1=3s⇒ usin⁡θg=3⇒usin⁡θ=30 and tan⁡β=usin⁡θ−gtucos⁡θ⇒tan⁡30∘=30−10×2ucos⁡θ⇒ ucos⁡θ=103 From here u=(103)2+302=203ms−1 and tan⁡θ=30103=3⇒θ=60∘

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