An airplane is flying at a height of $1960 m$ in the horizontal direction with a velocity of $360 k m / h r$ . It drops a bomb. The bomb strikes a point on the ground, then the time took $t$ by the bomb to reach the ground is:

20 \sqrt{2} sec

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20 sec

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10 \sqrt{2} sec

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10 sec

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Solution

The correct option is B $20 sec$
As the bomb is dropped so its initial vertical velocity will be $z e r o .$
so time taken by bomb will be given by $h = 0 + \frac{1}{2} g t^{2}$
or $t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 1960}{9.8}} = 20 s e c o n d$

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An airplane is flying at a height of 1960 m in the horizontal direction with a velocity of 360 km/hr. It drops a bomb. The bomb strikes a point on the ground, then the time took t by the bomb to reach the ground is:

The correct option is B 20secAs the bomb is dropped so its initial vertical velocity will be zero. so time taken by bomb will be given by h=0+12gt2or t=√2hg=√2×19609.8=20second

An airplane is flying at a height of $1960 m$ in the horizontal direction with a velocity of $360 k m / h r$ . It drops a bomb. The bomb strikes a point on the ground, then the time took $t$ by the bomb to reach the ground is:

The correct option is B $20 sec$
As the bomb is dropped so its initial vertical velocity will be $z e r o .$
so time taken by bomb will be given by $h = 0 + \frac{1}{2} g t^{2}$
or $t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 1960}{9.8}} = 20 s e c o n d$