Kinematics Question 631
Question: A cricket ball thrown across a field is at heights $ h _{1}, $ and $ h _{2} $ from point of projection at times $ t _{1} $ and $ t _{2} $ respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey it’s
Options:
A) $ \frac{h _{1}t _{2}^{2}-h _{2}t _{1}^{2}}{h _{1}t _{2}-h _{2}t _{1}} $
B)$ \frac{h _{1}t _{2}^{2}+h _{2}t _{1}^{2}}{h _{1}t _{2}+h _{2}t _{1}} $
C) $ \frac{h _{1}t _{2}}{h _{1}t _{2}-h _{2}t _{1}} $
D) None
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ {{\text{h}} _{\text{1}}}\text{=}\text{u sin}\theta {{\text{t}} _{\text{1}}}\text{+}\frac{\text{1}}{\text{2}}\text{gt} _{\text{1}}^{\text{2}}; $
$ {{h _{2}}}\text{=}\text{u sin}\theta {{t _{2}}}\text{+}\frac{\text{1}}{\text{2}}\text{gt} _{2}^{\text{2}} $
$ \text{So, }\frac{{{\text{t}} _{\text{1}}}}{{{t _{2}}}} =\frac{h _{1}+\frac{\text{1}}{\text{2}}\text{gt} _ {\text{1}}^{\text{2}}}{{{h _{2}}}+\frac{\text{1}}{\text{2}}\text{gt} _{2}^{\text{2}}} $
$ \Rightarrow h _{1}t _{2}-h _{2}t _{1}=\frac{1}{2}g( t _{1}t _{2}^{2}-t _{1}^{2}t _{2}^{{}} ) $
Time of flight $ \text{= 2u sin}\theta $
$\text{g=}\frac{h _{1}t _{2}^{2}-h _{2}t _{1}^{2}}{h _{1}t _{2}-h _{2}t _{1}} $ [Use above eqn. to simplify]
