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JEE MainPhysicsMotion in One Dimension

A hot air balloon is ascending vertically with a constant velocity. A first packet is dropped from the balloon and takes 9 s to reach the ground. A second packet is thrown vertically downwards relative to the balloon with a speed twice the balloon's ascent speed, and takes 4 s to reach the ground. If both packets are released at the same height, the height of the balloon at the instant the packets were released is __

Correct answer

180

Step-by-step solution

Let the upward velocity of the balloon be u . For the first packet dropped from the balloon, its initial velocity relative to the ground is u (upwards). The time taken to reach the ground is t₁ = 9 s . Using the second equation of motion: -h = u(9) - 1 2 g(9)^2 -h = 9u - 405 (1) For the second packet, it is thrown downwards relative to the balloon with speed 2u . Its initial velocity relative to the ground is u - 2u = -u (downwards). The time taken is t₂ = 4 s . -h = -u(4) - 1 2 g(4)^2 -h = -4u - 80 (2) Multiplying

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