Kinematics

JEE Main level, single correct, medium

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Question 151 of 308

A gas balloon is going up with a constant velocity of 1010 m/s. When this balloon reached a height of 7575 m, a stone is dropped from it and balloon keeps moving up with the same velocity. The height of the balloon when the stone hits the ground is ________ m. (Take g=10g=10 m/s2^2)

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Answer: (D) 125125

When the stone is dropped, it acquires the initial velocity of the balloon. Taking the upward direction as positive, the initial velocity of the stone is u=10u = 10 m/s.

The displacement of the stone when it hits the ground is S=−75S = -75 m, and its acceleration is a=−g=−10a = -g = -10 m/s2^2.

Using the equation of motion S=ut+12at2S = ut + \dfrac{1}{2}at^2:

−75=10t−12(10)t2-75 = 10t - \dfrac{1}{2}(10)t^2

−75=10t−5t2-75 = 10t - 5t^2

5t2−10t−75=05t^2 - 10t - 75 = 0

t2−2t−15=0t^2 - 2t - 15 = 0

(t−5)(t+3)=0(t - 5)(t + 3) = 0

Since time cannot be negative, t=5t = 5 s.

During this time, the balloon continues to move upwards with a constant velocity of 1010 m/s. The distance travelled by the balloon in 55 s is:

d=v×t=10×5=50d = v \times t = 10 \times 5 = 50 m

The total height of the balloon when the stone hits the ground is:

H=75+50=125H = 75 + 50 = 125 m

Answer: 125125

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