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 θ, 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 : cosθ
• D. 1 : secθ

Answer 1

Answer: (B)

1 : 1

Answer 2

Time of flight for the ball thrown by Pankaj $T_{1} = \frac{2u_{1}}{g}$

Time of flight for the ball thrown by Sudhir $T_{2} = \frac{2u_{2}\sin(90^{o} - \theta)}{g}$

$\frac { H _ { 1 } } { H _ { 2 } } = \left( \frac { T _ { 1 } } { T _ { 2 } } \right) ^ { 2 }$

According to problem $T_{1} = T_{2}$ ⇒ $\frac{2u_{1}}{g} = \frac{2u_{2}\cos\theta}{g}$ ⇒

$u_{1} = u_{2}\cos\theta$

Height of the ball thrown by Pankaj $H_{1} = \frac{u_{1}^{2}}{2g}$

Height of the ball thrown by Sudhir $H_{2} = \frac{u_{2}^{2}\sin^{2}(90^{o} - \theta)}{2g}$

$= \frac{u_{2}^{2}\cos^{2}\theta}{2g}$

∴ $\frac{H_{1}}{H_{2}} = \frac{u_{1}^{2}/2g}{u_{2}^{2}\cos^{2}\theta/2g}$ = 1 [As $u_{1} = u_{2}\cos\theta$]