a stone is released from the top of a tower of height h after its another stone is dropped from the balcony 20m below the top both reach the bottom simultaneously what is the value of h
Answers
Answer:
Explanation:
Let the time be t taken by first stone to reach ground from height h.
A/q
s=
2
1
gt
2
s=
2
1
10t
2
=5t
2
Similarly for second stone,
s=
2
1
gt
2
(h−20)=
2
1
10(t−1)
2
Solvig above two equations,
5t
2
−20=5(t
2
−2t+1)
⟹t=2.5s
now,
h=
2
1
10×2.5×2.5
2
=31.25m.
Answer:
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Explanation:
Let t seconds be the time taken for the first stone to reach the ground.
Using s = ut + 0.5at^2, with s = H, u = 0, a = 10:
H = 0.5 x 10 x t^2
H = 5t^2
The time taken for the second stone to reach the ground will be (t - 1) seconds. The stone will fall from a height of H - 20 metres.
Use s = ut + 0.5at^2, with s = H - 20, u = 0, a = 10, t = (t - 1):
H - 20 = 0.5 x 10 (t - 1)^2
H - 20 = 5(t - 1)^2
We know that H = 5t^2:
5t^2 - 20 = 5(t - 1)^2
5t^2 - 20 = 5(t^2 - 2t +1)
5t^2 - 20 = 5t^2 - 10t + 5
-20 = -10t +5
10t = 25
t = 2.5
Substituting for H:
H = 5 x 2.5^2 = 31.25 metres