Question

A body is thrown vertically upward with velocity u under earths gravity.what is the distance covered in last moment by the body?

Answer 1

Answer:

u² / 2 g

Explanation:

Given :

Initial velocity ( u ) = u m / sec

Acceleration due to gravity = - g   [ Acceleration against earth ]

Last moment mean final velocity ( v ) = 0

We have to find distance ( h ) :

We have equation of motion :

v² = u² - 2 g h

Putting values here we get :

0² = u² - 2 g h

2 g h = u²

h = u² / 2 g

Thus we get distance is  u² / 2 g  .

Answer 2

Initial velocity is u
At the highest the velocity of the body will be zero.
So by putting this value and cancelling negative signs on both sides of the equation formed by using the third equation of motion is
h=u^2/2g
or Distance or height till here is u square divided by twice of g(=9.8)
But at last moment the distance will be double because the body has returned to its original place. So
the total distance=2u^2/2g