Kinematics Question 631
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 is
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{{t_{1}}}{{t_{2}}}$ =$\frac{h {1}+\frac{\text{1}}{\text{2}}\text{gt}^2$ {\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{= } \frac{2u}{g} \text{ sin}\theta $
$\text{g=}\frac{h _{1}t _{2}^{2}-h _{2}t _{1}^{2}}{t _{2}-t _{1}} $ [Use the above equation to simplify]
BETA
