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
Given:
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.
To find:
If the two stones meet at just above the bottom of the building, find height of the building. (g = 10m/s^2)
Solution:
From given, we have,
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.
⇒ u = 30 m/s for second stone.
The distance covered by first stone after time t is s = 1/2gt²
The distance covered by second stone after time (t - 2) is s = u(t -2) + 1/2g(t - 2)² = 30t - 60 + 1/2g(t - 2)²
The two stones meet at just above the bottom of the building
⇒ 1/2gt² = 30t - 60 + 1/2g(t - 2)²
substituting the value of g = 10m/s², we get,
⇒ 1/2(10)t² = 30t - 60 + 1/2(10)(t - 2)²
⇒ 5t² = 30t - 60 + 5(t - 2)²
⇒ 5t² = 30t - 60 + 5(t² + 4 - 4t)
⇒ 5t² = 5t² + 10t - 40
⇒ 40 = 10t
solving the above quadratic equation, we get, t = 4
substitute the value t = 4 in s = 1/2gt², we get,
s = 1/2 × 10 × 4² = 80 m
Therefore, the height of the building is 80 m
Answer
80 m
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