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A balloon was moving upwards with a uniform velocity of $10 \mathrm{~m} / \mathrm{s}$ . An object of finite mass is dropped from the balloon when it was at a height of $75 \mathrm{~m}$ from the ground level. The height of the balloon from the ground when object strikes the ground was around:
(takes the value of $g$ as $10 \mathrm{~m} / \mathrm{s}^{2}$ )

Option: 1 $300 \mathrm{~m}$
Option: 2 $200 \mathrm{~m}$
Option: 3 $125 \mathrm{~m}$
Option: 4 $250\: m$

Answers (1)

best_answer

mass dropped will have same velocity as that of balloon

$s= ut+\frac{1}{2}at^{2}$
$-75= \left ( 10 \right )t+\frac{1}{2}\left ( -10 \right )t^{2}$
$-150=20t-10t^{2}$
$10t^{2}-20t-150= 0$
$t^{2}-2t-15= 0$
$\left ( t-5 \right )\left ( t+3 \right )= 0$
t = 5s
After 5s the mass will hit the ground
Height of the balloon at that instant will be
$H= \left ( 75 \right )+vt$
$= 75+\left ( 10 \right )\times 5$
$H=125 m$
The correct option is (3)

Posted by

vishal kumar

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