Free Falling body on Earth

FREE FALLING BODY ON EARTH

Consider a body of mass m to be dropped at a height of 1kilometer above the surface of Earth. The body will experience a force of gravity equal to the acceleration due to gravity of Earth g = 9.8 m/s² This value can be considered to be approximately same 1kilometer above Earth’s surface. As the body falls toward Earth, the displacement is considered negative.

ASSUMPTIONS

·         1. Effect of air resistance is negligible

·         2. Effect of wind is negligible

·         3. Effect of rotation of earth is negligible

Initial Displacement y­’ = 0 m

Final Displacement y = -1000 m

Applying first kinematic equation we get,

y - y' = u*t - (1/2)g*t^2

u – Initial velocity  (m/s)

-1000 = 0*t - (1/2)*9.8*t^2

-1000 = -4.9t^2

 t^2 = 204.08

 t = 14.2857 s

 

This is the time taken by the body to cover a distance of 1km.

The final velocity v when the body hits the surface is given by 3^rd Kinematic equation,

v^2 = u^2 - 2g(y - y')

v^2 = 0 - 2*9.8*(-1000 - 0)

                                                             

v^2 = 2*9.8*1000

v = 140 m/s

Thus the final velocity of the body when it reaches surface of Earth is 140 m/s or 504kmph or 315mph