5.11 A truck starts from rest and accelerates uniformly at 2.0 m s-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s? (Neglect air resistance.)
5.11 A truck starts from rest and accelerates uniformly at 2.0 m s-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s? (Neglect air resistance.)
The acceleration of the truck, a = 2 m/s2, t = 10 s, Initial velocity, u = 0, from the equation v = u + at, we get v = 2 = 20 m/s
At time t = 11 s, the horizontal component of the velocity Vx remains unchanged due to no air resistance. Hence Vx = 20 m/s
The vertical component of the velocity Vy is ex
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= 0.92 * 1260 = 1161 m/s
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T – 2g sin37 = 2a . (i)
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4g – 2T =
2g – T = a . (ii)
T = (2g – a)
2g – a – 2g × = 2a
3a = 2g ×
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physics ncert solutions class 11th 2023
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