Question

A ball is thrown up with a velocity of 0.5 m/s .how high will it go before it begins to fall ?

Answer 1

given :-
   u = 0.5 m/s
   g = 9.8 m/s^2 ( it is taken to be negative since body is going against the gravity .
   v = 0
   s = ?

v^2 - u^2 = 2gs 
(0)^2 - (0.5)^2 = 2(-9.8) S
- (0.5 * 0.5 ) =  -2  * 9.8 * S
              S = 0.5 *0.5 / 2 * 9.8 = 0.012 m is answer


  

Answer 2

Initial  velocity u = 0.5 m/sec

Final velocity v = 0 m/sec

acceleration due to gravity g = 9.8m/sec²

v² - u² = 2gH

0² - (0.5)² = 2 * (-9.8)*H

H = 0.012 m