A body is dropped from height h. it covers 9/25 part of distance in last sec of it's motion find the height and time of flight.
Answers
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)
PS:
Alternatively (3) looks like
4^2 * t^2 = 5^2 * (t-1)^2
==> (5–1)^2 * t^2 = (t-1)^2 * 5^2
Good Luck
Answer:
During its motion it covers 925 part of height of tower in the last 1 second Then find the height of tower. ⇒n=50+√502-4×9×(+25)2×9=5sec.
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