Question

From a certain height, a helicopter starts going up with acceleration. In 10 sec, it goes up by 1/6th of its original height. If a ball is dropped at this instant, it takes 5 sec to reach the ground. What is the original height from which the helicopter has started and with what acceleration?

Answer 1

Need the complete solution for this. :(

Answer 2

The formula for distance (d), for constant acceleration (a) and time t:

d==(1/2) a t^2. For gravity a = g = 9.8 m/s2, call it 10. In 5 s, we have

d= (1/2)(10)(25) = 125 m.

d is 1 + 1/6 of original distance, D, d = 125m = (6/5) D

D = (5/6) d= 104 m, original distance.

Answer 3

THe answers are h=91.875m and a=.30625m/s^2

Answer 4

I see. Sorry, (1+ 1/6)=7/6, and I should have had 125m=(7/6)D,
D=(6/7)125=107m.

I am still puzzled. Certainly 5 sec means (1/2)gt^2=(0.5)(10)(25)=125m.
Was the stone dropped before the acceleration or after? "At this instant" is not clear.

Once we have the original height, we can get the additional height h' and use it to calculate the acceleration a from h' = (1/2) a t^2.