A ball of mass $100$ gm is projected vertically upwards from the ground with a velocity of $49$ $m / s$ . At the same time another identical ball is dropped from a height of $98$ m. After some time the two bodies collide. When they collide, their velocities are (Given $g = 9.8 m / s^{2}$ )

29.4 m/s upwards, 29.4 m/s downwards

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29.4 m/s upwards, 19.6 m/s downwards

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19.6 m/s upwards, 19.6 m/s downwards

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Solution

The correct option is B 29.4 m/s upwards, 19.6 m/s downwards
Equations to use :
time of meet $t = \frac{h}{u}$
$v_{1} = g t$ ( for freely falling body)
$v_{2} = u - g t$ ( for body projected up)

For A: $s_{A} = 49 t - \frac{1}{2} g t^{2} - (1)$
For B: $s_{B} = \frac{1}{2} g t^{2} - (2)$
$s_{A} + s_{B} = 98 = 49 t (f r o m (1) & (2))$
$\Rightarrow t = 2 s e c$
So, $v_{A} = u_{A} - g t = 49 - 9.8 \times 2 = 29.4 m / s$ upwards
$v_{B} = g t = 9.8 \times 2 = 19.6 m / s$ downwards

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A ball of mass 100 gm is projected vertically upwards from the ground with a velocity of 49 m/s. At the same time another identical ball is dropped from a height of 98 m. After some time the two bodies collide. When they collide, their velocities are (Given g=9.8m/s2)

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