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)} $
BETA
