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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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