Question
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/s2 )
15 m
10 m
20 m
5 m
Solution
B.
10 m
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
v2 = u2 - 2gh
v2 - final velocity ; u2 = initial velocity
At A,
⇒ u2 = 100 + 10 H ....(i)
At 'O'
( 0)2 = u2 - 2 × 10 × H ....(ii)
Thus, from Eqs. (i) and (ii), we get
20 H = 100 + 10 H
⇒ 10 H = 100
∴ H = 10 m
