Kinematics Question 398

Question: A body is projected vertically upwards. If $ t _{1} $ and $ t _{2} $ be the times at whichis at height h above the projection while ascending and descending respectively, then h it’s

Options:

A) $ \frac{1}{2}gt _{1}t _{2} $

B) $ gt _{1}t _{2} $

C) $ 2gt _{1}t _{2} $

D) $ 2hg $

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Answer:

Correct Answer: A

Solution:

$ h=ut _{1}-\frac{1}{2}gt _{1}^{2} $

$ \text{Also h=u}{{t} _{2}}-\frac{1}{2}gt _{2}^{2} $

After simplify above equations, we get $ h=\frac{1}{2}gt _{1}{t _{2.}} $



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