a body is droped from a heigt of 250m and semutaneously another body is droped from a height of 200m what is a diffrencein height after they have fallen for 5 seconds{g=9.8ms-2}
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Answer:
hi here is your answer dear friend
Explanation:
The difference in height after they have fallen for 5 seconds is 50m.
Explanation:
Given,
A body is dropped from a height of 250m and simultaneously another body is dropped from a height of 200m.
∴ Initially, the difference between their heights is = 250 - 200 = 50m.
Now, we've to calculate the difference between their heights after they've fallen for 5 sec.
Let's calculate the distance covered by the first body in 5 sec.
We know that,
→ s = ut + ½.g.t²
Where,
u(intial velocity) = 0 [Since it was at rest]t(time) = 5sec.g(acceleration due to gravity) = 9.8m/s²
Substitute the values,
→ s = 0(5) + ½.(9.8)(5)²
→ s = 0 + ½.(9.8)(25)
→ s = 122.5m
∴ Remaining distance = 250 - 122.5 = 127.5m
Similarly, for the second object :-
→ s = ut + ½.a.t²
→ s = 0 + ½.(9.8)(25)
→ s = 122.5m
∴ Remaining distance = 200 - 122.5 = 77.5m
Hence, after falling for 5 seconds the distance between them would be :-
= 127.5 - 77.5
= 50m
∴ Required answer:- 50m
______________________
Formula used,
s = ut + ½at²
Where,
s denotes the distance.
u = intial velocity.
v = final velocity.
t = time taken.
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