A stone is released from top of a building. After 2 seconds, another stone is projected vertically downwards along same line with speed of 30 m/s. If the two stones meet at just above the bottom of the building, find height of the building. (g = 10m/s^2)
Answers
For upper stone:
u1 = 0
x = 0 + 1/2 gt*2
x = 1/2 x 10 x t*2
x = 5t*2 – (1)
Height of building is 80 m
Explanation:
A stone is released from top of a building
=> u = 0
Let say it reached bottom of building after t sec
S = ut + (1/2)at²
S = height of building = h m
a = g = 10 m/s²
=> h = 0 + (1/2)(10)t²
=> h = 5t²
After 2 seconds, another stone is projected vertically downwards along same line with speed of 30 m/s
=> u = 30
will reach at bottom after t - 2 secs
h = 30*(t - 2) + (1/2)(10)(t - 2)²
=> h = 30t - 60 + 5 (t - 2)²
=> h = 30t - 60 + 5 ( t² - 4t + 4)
=> h = 30t - 60 + 5t² - 20t + 20
=> h = 5t² + 10t - 40
Equating h
5t² = 5t² + 10t - 40
=> 10t = 40
=> t = 4
h = 5t² = 5(4)² = 80 m
Height of building is 80 m
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