Kinematics Question 514

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 $

Show Answer

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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