Question

a body released from certain height above the ground describes 9/25 of the height in the last second of its fall then it is falling from a height equal to​

Answer 1

Answer:

125 metres

Explanation:

Ball travels 9/25 of the height (say x) in the t second(last second)

Hence it travelled 16/25 x in (t-1) seconds.

and the total distance in t seconds.

The ball is dropped hence initial velocity is zero and gravity is 9.8m/s^2 or g

Hence distance travelled is

s = 1/2 * a * t^2 = (1/2) g t^2

==> x = (1/2) gt^2 ______(1)

And 16x/25 = (1/2)g(t-1)^2 _______(2)

Divide the equations (2) by (1)

16/25 = (t-1)^2/t^2

==> 16t^2 = 25(t-1)^2 _____(3)

Which gives t = 5 seconds (Time cannot be negative, atleast not in this case)

Hence height x = (1/2)*9.8*(5^2) = 122.5 meters

or 125 meters (if you take g=10m/s^2)