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Q.

At what angle should a body be projected with a velocity 24 ms-1 just to pass over the obstacle 14 m high at a distance of 24 m? [Take, g =10 ms-2]

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a

tan−1⁡(3.8)

b

tan−1⁡ (1)

c

tan−1⁡ (3.2)

d

tan−1⁡(2)

answer is A.

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Detailed Solution

y=xtan⁡θ−gx22u2cos2⁡θ⇒14=24tan⁡θ−5(24)2(24)2sec2⁡θ⇒5tan2⁡θ−24tan⁡θ+19=0⇒tan⁡θ=1 and tan⁡θ=195=3.8⇒θ=tan−1⁡(1) and θ=tan−1⁡(3.8)
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