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

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