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An aeroplane is moving with horizontal velocity u at height h. The velocity of a packet dropped from it on the earth's surface will be (g is acceleration due to gravity)
- a)√u2+2gh
- b)√2gh
- c)2 gh
- d)√u2-2gh
Correct answer is option 'A'. Can you explain this answer?
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An aeroplane is moving with horizontal velocity u at height h. The vel...
U
b) u + g
c) u - g
d) g
Answer:
c) u - g
Explanation:
When the packet is dropped from the aeroplane, it will have the same horizontal velocity as the aeroplane, which is u. However, it will also be subjected to the acceleration due to gravity, which is downwards. This acceleration will cause the packet to gain a vertical velocity component. The velocity of the packet at any instant can be resolved into two components: the horizontal component (u) and the vertical component (v). The vertical component will increase with time due to gravity, and the horizontal component will remain constant. Therefore, the velocity of the packet at any instant will be the vector sum of these two components. At the instant when the packet reaches the ground, its vertical velocity component will be equal to the velocity that it would have acquired due to gravity if it had been dropped from rest on the ground. This velocity is given by v = gt, where t is the time taken to reach the ground. Therefore, the velocity of the packet just before it hits the ground will be (u, -gt). The magnitude of this velocity is given by (u^2 + (-gt)^2)^0.5, which is equal to (u^2 + g^2t^2)^0.5. As t is the time taken to reach the ground, it is given by t = (2h/g)^0.5 (by using the formula h = 1/2gt^2). Therefore, the magnitude of the velocity of the packet just before it hits the ground is (u^2 + 2gh)^0.5. The direction of this velocity is downwards, so the answer is u - g.
b) u + g
c) u - g
d) g
Answer:
c) u - g
Explanation:
When the packet is dropped from the aeroplane, it will have the same horizontal velocity as the aeroplane, which is u. However, it will also be subjected to the acceleration due to gravity, which is downwards. This acceleration will cause the packet to gain a vertical velocity component. The velocity of the packet at any instant can be resolved into two components: the horizontal component (u) and the vertical component (v). The vertical component will increase with time due to gravity, and the horizontal component will remain constant. Therefore, the velocity of the packet at any instant will be the vector sum of these two components. At the instant when the packet reaches the ground, its vertical velocity component will be equal to the velocity that it would have acquired due to gravity if it had been dropped from rest on the ground. This velocity is given by v = gt, where t is the time taken to reach the ground. Therefore, the velocity of the packet just before it hits the ground will be (u, -gt). The magnitude of this velocity is given by (u^2 + (-gt)^2)^0.5, which is equal to (u^2 + g^2t^2)^0.5. As t is the time taken to reach the ground, it is given by t = (2h/g)^0.5 (by using the formula h = 1/2gt^2). Therefore, the magnitude of the velocity of the packet just before it hits the ground is (u^2 + 2gh)^0.5. The direction of this velocity is downwards, so the answer is u - g.
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An aeroplane is moving with horizontal velocity u at height h. The velocity of a packet dropped from it on the earths surface will be (g is acceleration due to gravity)a)√u2+2ghb)√2ghc)2 ghd)√u2-2ghCorrect answer is option 'A'. Can you explain this answer?
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An aeroplane is moving with horizontal velocity u at height h. The velocity of a packet dropped from it on the earths surface will be (g is acceleration due to gravity)a)√u2+2ghb)√2ghc)2 ghd)√u2-2ghCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An aeroplane is moving with horizontal velocity u at height h. The velocity of a packet dropped from it on the earths surface will be (g is acceleration due to gravity)a)√u2+2ghb)√2ghc)2 ghd)√u2-2ghCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aeroplane is moving with horizontal velocity u at height h. The velocity of a packet dropped from it on the earths surface will be (g is acceleration due to gravity)a)√u2+2ghb)√2ghc)2 ghd)√u2-2ghCorrect answer is option 'A'. Can you explain this answer?.
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