A body A is projected upwards with velocity $v_{1}$ . Another body B of same mass is projected at an angle of $45^{0}$ with the horizontal. Both reach the same height. What is the ratio of their initial kinetic energies?

1/4

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1/3

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1/2

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Solution

The correct option is B 1/2
Let the velocity of body $A$ is $v_{1}$ and mass is $m$
Let Velocity of body $B$ is $v_{2}$ and mass is $m$ and it is projected at angle of $45^{o}$ with the horizontal$

Maximum height reached by $A, h_{1} = \frac{v_{1}^{2}}{2 g}$

Maximum height reached by $B, h_{2} = \frac{v_{2}^{2} s i n^{2} 45^{o}}{2 g}$

Given that $h_{1} = h_{2}$

$⟹ \frac{v_{1}^{2}}{2 g} = \frac{v_{2}^{2} s i n^{2} 45^{o}}{2 g}$

$⟹ \frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{2}$

Hence Required ratio of their initial kinetic energy, $\frac{K . E ._{1}}{K . E ._{2}} = \frac{\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v_{2}^{2}} = \frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{2}$

Hence option $(C)$ is correct.

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A body A is projected upwards with velocity v1. Another body B of same mass is projected at an angle of 450 with the horizontal. Both reach the same height. What is the ratio of their initial kinetic energies?

A body A is projected upwards with velocity $v_{1}$ . Another body B of same mass is projected at an angle of $45^{0}$ with the horizontal. Both reach the same height. What is the ratio of their initial kinetic energies?