question_answer

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

A) 2 : 1                

B) 1 : 1

C) \[1:cos\theta \]  

D) \[1:sec\theta \]

Correct Answer: B

Solution :

[b] Time of flight for the ball thrown by pankaj, \[{{T}_{1}}=\frac{2{{u}_{1}}}{g}\]time of flight for the ball thrown by sudhir \[{{T}_{2}}=\frac{2{{u}^{2}}\sin (90{}^\circ -\theta )}{g}-\frac{2{{u}^{2}}\cos \theta }{g}\] According to problem \[{{T}_{1}}={{T}_{2}}\] \[\Rightarrow \frac{2{{u}_{1}}}{g}=\frac{2{{u}^{2}}\cos \theta }{g}\] \[\Rightarrow {{u}_{1}}={{u}_{2}}\cos \theta \] Height of the ball thrown by pankaj \[{{H}_{1}}=\frac{{{u}_{1}}^{2}}{2g}\] Height of the thrown by sudhir \[{{H}_{2}}=\frac{{{u}_{2}}^{2}{{\sin }^{2}}(90{}^\circ -\theta )}{2g}\]\[{{H}_{2}}=\frac{{{u}_{2}}^{2}{{\sin }^{2}}(90{}^\circ -\theta )}{2g}=\frac{{{u}_{2}}^{2}{{\cos }^{2}}\theta }{2g}\] \[\therefore \frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{u}_{1}}^{2}/2g}{{{u}_{2}}^{2}{{\cos }^{2}}\theta /2g}=1\]\[[As{{u}_{1}}={{u}_{2}}cos\theta ]\]