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Standard XII

Physics

Motion Under Gravity

Two balls are...

Question

Two balls are dropped from different heights at different instants. Second ball is dropped $2$ sec after the first ball. If both balls reach the ground simutlanously after $5$ sec of dropping the first ball, the difference of initial heights of the two balls will be: $(g = 9.8 m / s^{2})$

58.8 m

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78.4 m

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98.0 m

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117.6 m

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Solution

The correct option is A $78.4 m$
$h_{1} - h_{2} = \frac{1}{2} g (t_{1}^{2} - t_{2}^{2})$
$= \frac{1}{2} \times 9.8 [(5)^{2} - (3)^{2}]$
$= 8 \times 9.8$ $= 78.4 m$

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Two balls are dropped from different heights at different instants. Second ball is dropped 2 sec after the first ball. If both balls reach the ground simutlanously after 5 sec of dropping the first ball, the difference of initial heights of the two balls will be: (g=9.8 m/s2)

The correct option is A 78.4mh1−h2=12g(t21−t22)=12×9.8[(5)2−(3)2]=8×9.8 =78.4m

Two balls are dropped from different heights at different instants. Second ball is dropped $2$ sec after the first ball. If both balls reach the ground simutlanously after $5$ sec of dropping the first ball, the difference of initial heights of the two balls will be: $(g = 9.8 m / s^{2})$

The correct option is A $78.4 m$
$h_{1} - h_{2} = \frac{1}{2} g (t_{1}^{2} - t_{2}^{2})$
$= \frac{1}{2} \times 9.8 [(5)^{2} - (3)^{2}]$
$= 8 \times 9.8$ $= 78.4 m$