First slide

Projectile motion

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 $θ w i t h t h e v e r t i c l e$ , both of them stay
in our view for the same period. The height attained by the two balls are in the ratio

Difficult

A

2:1
B

1:1
C

$1 : \cos θ$
D

$1 : \sec θ$

Solution

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
$T_{2} = \frac{2 u_{2} \sin (90^{\circ} - θ)}{g} = \frac{2 {u_{2}}^{} \cos θ}{g}$
According to problem $T_{1} = T_{2}$
$\begin{array}{l} \Rightarrow \frac{2 u_{1}}{g} = \frac{2 u_{2} \cos θ}{g} \\ \Rightarrow u_{1} = u_{2} \cos θ \end{array}$
Height of the ball thrown by pankaj $H_{1} = \frac{u_{1}^{2}}{2 g}$
Height of the ball thrown by Sudhir
$\begin{array}{l} H_{2} = \frac{u_{2}^{2} \sin^{2} (90^{\circ} - θ)}{2 g} = \frac{u_{2}^{2} \cos^{2} θ}{2 g} \\ ∴ \frac{H_{1}}{H_{2}} = \frac{u_{1}^{2} / 2 g}{u_{2}^{2} \cos^{2} θ / 2 g} = 1 [As u_{1} = u_{2} \cos θ] \end{array}$

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