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A lift starts from rest at t = 0 and moves upwards with constant acceleration of 5 m/s2. At t = 2 s, a boltfrom the ceiling of lift detaches and falls freely. If height of lift is 7.5 m, then the bolt hits the floor of lift at1= ns. Find the value of n. (g = 10 m/s2) |
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Answer» Answer: Explanation: A lift starts from rest at t = 0 and moves upwards with constant acceleration of 5 m/s2. At t = 2 s, a bolt from the ceiling of lift DETACHES and falls freely. If height of lift is 7.5 m, then the bolt hits the FLOOR of lift at 1= ns. Find the value of n. (G = 10 m/s2) Lift starts from rest u = 0 m/s after t = 2sec a = 5 m/s² v = u + at = 0 + 5*2 = 10 m/s Bolt hits the floor in n sec Bolt initial speed = 0 m/s a = 10 m/s² S = ut + (1/2)at² => S bolt = 0 + (1/2)(10)n² => S bolt = 5n² SLift = 10n + (1/2)5n² => SLift = 10n + 2.5n² S bolt + Slift = 7.5 => 5n² + 10n + 2.5n² = 7.5 => 7.5n² + 10n - 7.5 = 0 => 3n² + 4n - 3 = 0 => n = 0.535 sec |
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