Question

a body is thrown vertically up with a speed of 100 m/s.on the return journey,the speed in m/s at the starting point will be 
a)100 m/s  b)9.8 m/s c)100×9.8 m/s d)100÷9.8 m/s

The answer in my textbook is a) 100 m/s.

Please solve it urgently.

Answer 1

it is same 100 m/sec

v = final velocity     u = initial velocity     a = acceleration 
S = distance traveled      t = time traveled
suppose at the top point it reaches, its height is H.

A body goes up to a height during the upward journey
            v = u + at      =>  0 = u - g t        =>  t = u/g  : time to reach the top
            v² - u² = 2 a S    =>  0² - u² = 2 (-g) H  => u² = 2 g H    => H = 1/2u²/g

OR,          H = ut + 1/2 a t²  =  u*u/g -1/2 g (u/g)² = u²/g - 1/2 u²/g

       So H = 1/2 u² / g    ------ equation 1

A body falls freely from height H: then
           v = u+at    =>    v = 0+ g t  = gt
           v² - u² = 2 a S  =>  v² - 0 = 2 g S      =>  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                --- which is same as equation 1

So velocity at a height H from ground on the way up
           = velocity at same height H from ground on the way down