1. A stone is dropped freely on the floor from a building. It takes 3 secs to reach the ground. Calculate the height of the building. Take the normal value of g.
Answers
Answer:
The distance an object moves, starting from rest, under a constant acceleration is equal to (1/2)at^2.* Gravitational acceleration changes with height; infinitely far from Earth, the acceleration is zero. Close to the Earth’s surface, the acceleration is a constant 9.81 meters per second per second.
But for any building built so far in history, the change in acceleration between top and bottom can be taken to be entirely negligible. This means that, in the case you are asking about, we can assume a = 9.81 meters per second per second.
So, let's take that 10 seconds and solve for the height. We have:
h = (1/2) a t^2 = (1/2) (9.81) (10)^2 = 490.5 meters
The building is 490.5 meters tall!
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*If you want to confirm that for yourself, you'll have to solve the differential equation (d/dt)^2 y = a, where a is a constant. If you don't know what that means, don't worry; based on the fact that you are asking this question, deriving this formula likely involves mathematics higher than you've learned.