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Question

A ball is projected obliquely with a velocity 49 ms1 strikes the ground at a distance of 245 m from the point of projection. It remained in air for:

A
10 sec
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B
52 sec
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C
3 sec
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D
2.5 sec
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Solution

The correct option is B 52 sec
The ball is projected obliquely so suppose of projection from horizontal was θ
Then the horizontal range should be R=u2Sin2θg
As given in the question R=245m and u=49m/s
putting above values in the expression of R we get Sin2θ=1 so 2θ=900 or θ=450 so Sinθ=122
So the time in air or the time of flight will be T=2uSinθg=2×49Sinθ9.8=522Sec
Option B is correct.

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