An elevator is descending with uniform acceleration.To measure the acceleration,a person in the elevator drops a coin at the moment the elevator starts.The coin is 6 ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second.Calculate from these data the acceleration of the elevator.

Question

Pavan kumar · Answer

Let acceleration of lift be a.

acceleration of ball with respect to the lift = acceleration of ball with respesct to ground - acceleration of the lift

Or, a_ball(lift) = g - a = 9.8 m/s^2 - a = 32 ft./s^2 - a [ 9.8m = 32 ft. ]

displacement s = 6 ft.

time taken t = 1 s.

initial velocity u = 0

s = ut + 1/2 at^2 = 0*1 + 1/2 (32-a)*1*1 = 1/2 (32-a)

Therefore, 6 = 1/2 (32-a)

Hence a = 20 ft./s^2

Ashish Ranjan · Answer

The above answer for the question ‘‘ An elevator is descending ...’‘ is absolutely correct i just wanted to clear some concepts . That is, Here the term “acceleration of coin with respect to the lift “ is used beacuse the observer is present in the lift,so the accleration of coin that he would observe is “the accleration of coin w.r.t lift , if the observer is present outside the lift , we would simply take into account the “ACTUAL ACCLERATION OF THE COIN”. Second Thing that i want to conclude that - a person in the elevator drops a coin at the moment the elevator starts. that means the initial velocity of the coin would also be 0.So, Don’t confuse with this term - The coin is dropped at the instant when evrything is at standstill . Acceleration = (2d/t^2), = 6.142m/sec^2. 6ft. is 1.829m., so distance elevator has drolled in the 1 sec = (4.9 - 1.829) = 3.071m. Height of fall in 1s = 1/2 (t^2 x g), = 4.9m. Thsi should be the answer.. a=6.14m/sec^2 3.07=0+1/2*a*1 S=ut+1/2at^2 (where a is the acceleration of elevator) So substituting this S again we have but the distance to floor was only 1.82m(6ft) that means the elevator moved 4.9-1.82=3.07m in that one second. S=0x1+1/2x9.8x1=4.9 m S=ut+1/2gt^2 the coin drops after one sec that means distance covered by coin

Dweepayan Biswal · Answer

20 check out hc verma.?........................................................?...........?.?..................?.................................................?

navya · Answer

for elevator and coin u=0 acceleration of the coin w.r.t to earth – acceleeration of the lift is basicaly = g-a ( it is coin -earth) = 9.8- a height of the coin from the elevator is 6ft = 1.8 m aprox t =1s s= ut+1/2at 2 1.8 = 0t = ½ (9.8 -a) 1 2 = 6.2 m/s 2

Diya Biju · Answer

S=6ft g=32ft/s^2 t=1 s u=0 So, s=ut+1/2at^2 6=1/2×a a=12ft/s^2 (By the relative velocity concept) Acceleration of the elevator=32_12=20ft/s^2

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