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`        A lift is descending  with uniform acc. To measure the acc a person in the lift drops a coin at the moment  at the moment when lift was descending with speed 6 ft / s. The coin is 5ft above the floor of the lift when it was dropped. The person observes that the coin strikes the floor in 1 sec. calculate the acc of the lift.`
one year ago

Khimraj
3008 Points
```							cceleration due to gravity = 9.8 m/s² = 32 ft/s².∵ Acceleration of the the Lift is equals to the = Acceleration of Coin w.r.t the Ground - Acceleration of the Coin w.r.t. the lift.∴ Acceleration of the Coin w.r.t. the lift = Acceleration of the Coin w.r.t. the Ground - Acceleration of the Lift.∴ Acceleration of the Coin w.r.t. the lift = (g - a) ft/s².       = (32 - a)ft/s²   From the Question, Time taken by the Coin to strikes the Ground (t) = 1 seconds.Distance covered through the coin before striking the ground (S) = 5 ft. Initial Velocity(u) of the coin w.r.t. the lift = 0[Since, the coin starts from the rest.]From the Second Equation of the Motion,S = ut + (1/2)a₁t².Where,a₁ = acceleration of the Coin w.r.t. the lift.5 = 0 + (1/2)a₁(1)² 5 = 0 + a₁/2 a₁ =  10 a₁ = 10  ft/s².Now, Acceleration of the Coin w.r.t. the lift = 10 ft/s².∴  10 = 32 - a   a = 32 - 10   a =  22 ft/s².∴ Acceleration of the lift is 22 ft/s².Hope it helps.
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one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions