A ball is thrown vertically upward. It has a speed of 10 ~m / s when it has reached one-half of its maximum height. How high does the ball rise? (Take, g=10 ~m / s ^2 )

Step-by-step solution

The problem can be solved using third equation of motion at A and O '. Let maximum height attained by the ball be H.Third equation of motion gives v^2=u^2-2 g h At A , (10)^2=u^2-2 10 H 2 u^2=100+10 H...(i) At O^ ,(0)^2=u^2-2 10 H u^2=20 H...(ii) Thus, from Eqs. (i) and (ii), we get 20 H=100+10 H 10 H=100 H=10 ~m