Kinematics Question 412
Question: A ball is thrown vertically upwards. It was observed, at a height h twice with a time interval $ \Delta t $ . The initial velocity of the ball is
Options:
A) $ \sqrt{8gh+g^{2}{{( \Delta t )}^{2}}} $
B) $ \sqrt{8gh+{{( \frac{g\Delta t}{2} )}^{2}}} $
C) $ \frac{1}{2}\sqrt{8gh+g^{2}{{( \Delta t )}^{2}}} $
D) $ \sqrt{8gh+4g^{2}{{( \Delta t )}^{2}}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \text{h=ut}-\frac{1}{2}a{{t}^{2}} $
$ \Rightarrow gt^{2}-2ut+2h=0 $
Solving for t we get $ t _{1}+t _{2}=2u/g $
$ t _{1}\times t _{2}=2h\text{/}g $
$ \text{so, }\Delta \text{t= }|t _{1}-t _{2}|\text{ =}{{( t _{1}+t _{2} )}^{2}}-4t _{1}t _{2} $
Putting value we get $ \text{u=}\frac{1}{2}\sqrt{8hg+g^{2}\Delta t^{2}} $
BETA
