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 $
Show Answer
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.}} $
BETA
