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) I NEED ANSWER FAST PLS FOR THAT I AM KEEPING THIS QUESTION FOR 50 POINTS
Answers
Solution:
Speed of first stone, u = 0 m/s
Acceleration , a = g(Acceleration due to gravity)
Distance covered by first stone after time t =>
Apply Equation of Motion
S = ut + (1/2)at²
S = 0t + (1/2)gt²
S = (1/2)gt² [Equation 1]
Speed of second stone, u = 30 m/s
Acceleration , a = g(Acceleration due to gravity)
Distance covered by second stone after time (t - 2) =>
Apply Equation of Motion
S = ut + (1/2)at²
S = u(t - 2) + 1/2g(t - 2)²
S = 30t - 60 + 1/2g(t - 2)² [Equation 2]
Now, it is said in the question if the two stones meet just above the bottom of the building.
Equating Equation 1 and Equation 2
1/2gt² = 30t - 60 + 1/2g(t - 2)²
Put value of g = 10 m/s²
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 = 30t - 60 + 5t² + 20 - 20t
5t² - 5t² = 10t - 40
10t - 40 = 0
t = 40/10
t = 4 seconds
Put value of t in Equation 1
S = (1/2)gt²
S = (1/2)(10)(4)²
S = 5 * 16
S = 80 m
The height of the building is 80 m.
Answer:
height of the building is 80m