Physics, asked by Jay1411, 8 months ago

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

Answered by BrainlyYoda
18

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.

Answered by LEGENDARYSUMIT01
0

Answer:

height of the building is 80m

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