Question

A body is thrown vertically upwards with a speed of 100 m/s on the return journey speed in m/s at the starting point will be

Answer 1

v = 100 m/s

returning point .. as it attend its maximum height ... so its velocity become zero ..

now when starting point .. its veloctiy is 0 , but due to accleration velocity will increase .

Answer 2

Answer: 100 m/s

Suppose:

At the top point it reaches, its height is H.

And:

A body goes up to a height during the upward journey

$\boxed{\sf{v = u + at}}$

$\implies$ 0 = u - g t       

$\implies$ t = u/g 

And:

$\boxed{\sf{v^{2} - u^{2} = 2\:aS}}$

$\implies$ 0² - u² = 2 (-g) H 

$\implies$ u² = 2 g H   

$\implies$ H = 1/2 u²/g

So: H = 1/2 u² / g  ---- eq. (1)

A body falls freely from height H:

Then:

$\boxed{\sf{v^{2} - u^{2} = 2\:aS}}$

$\implies$ v² - 0 = 2 g S     

$\implies$ S = 1/2 v² / g

If the body travels a distance H on the way down then, above formula is:              

H = 1/2 v² / g               

Note: (Which is same as equation 1)

So:

Velocity at a height H from ground:

$\boxed{\sf{On\:the\:way\:up = On\:the\:way\:down}}$