Kinematics Question 622

Question: A particle is projected at an angle of elevation $ \alpha $ and after t seconds it appears to have an angle of elevation $ \beta $ as seen from point of projection. The initial velocity will be

Options:

A) $ \frac{gt}{2\sin ( \alpha -\beta)} $

B)$ \frac{gt\cos \beta }{2\sin ( \alpha -\beta)} $

C) $ \frac{\sin ( \alpha -\beta)}{2gt} $

D)$ \frac{2\sin ( \alpha -\beta)}{gt\cos \beta } $

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Answer:

Correct Answer: B

Solution:

[b] $ t=\frac{2u\sin ( \alpha -\beta)}{g\cos \beta }.\text{};u=\frac{gt\cos \beta }{2\sin ( \alpha -\beta)} $



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