Physics Two Dimensional Motion Question 61
Question: Pankaj and Sudhir are playing with two different balls of masses m and $ 2m $ , respectively. If Pankaj throws his ball vertically up and Sudhir at an angle$ \theta $ , both of them stay in our view for the same period. The height attained by the two balls are in the ratio
Options:
A) 2 : 1 B) 1 : 1 C) $ 1:cos\theta $ D) $ 1:sec\theta $
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Answer:
Correct Answer: B
Solution:
[b] Time of flight for the ball thrown by pankaj, $ T _{1}=\frac{2u _{1}}{g} $ time of flight for the ball thrown by sudhir $ T _{2}=\frac{2u^{2}\sin (90{}^\circ -\theta )}{g}-\frac{2u^{2}\cos \theta }{g} $ According to problem $ T _{1}=T _{2} $
$ \Rightarrow \frac{2u _{1}}{g}=\frac{2u^{2}\cos \theta }{g} $
$ \Rightarrow u _{1}=u _{2}\cos \theta $ Height of the ball thrown by pankaj $ H _{1}=\frac{u _{1}^{2}}{2g} $ Height of the thrown by sudhir $ H _{2}=\frac{u _{2}^{2}{{\sin }^{2}}(90{}^\circ -\theta )}{2g} $
$ H _{2}=\frac{u _{2}^{2}{{\sin }^{2}}(90{}^\circ -\theta )}{2g}=\frac{u _{2}^{2}{{\cos }^{2}}\theta }{2g} $
$ \
Therefore \frac{H _{1}}{H _{2}}=\frac{u _{1}^{2}/2g}{u _{2}^{2}{{\cos }^{2}}\theta /2g}=1 $
$ [Asu _{1}=u _{2}cos\theta ] $
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