A uniform solid spherical ball is rolling down a smooth inclined plane from a height h. The velocity attained by the ball when it reaches the bottom of the inclined plane is v. If the ball is now thrown vertically upwards with the same velocity v, the maximum height to which the ball will rise is

Options

A. 5h/8
B. 3h/5
C. 5h/7
D. 7h/9

Answer

C

Explanation

Let m be the mass and R be the radius of the ball . According to law of conservation of mechanical energy

`mgh = (1)/(2) mv^(2) + (1)/(2) I omega^(2)`
`= (1)/(2) mv^(2) + (1)/(2) ((2)/(5) m R^(2)) omega^(2)`
(for solid sphere , `I = (2)/(5) mR^(2))`
` = (1)/(2) mv^(2) + (1)/(5) mv^(2) (because v = R omega)`
`mgh = (7)/(10) mv^(2)` or `v = sqrt((10)/(7) gh)`
Let h. be maximum height attained by the ball when it is thrown vertically upwards with velocity `v ( = sqrt((10)/(7) gh))`
`therefore h. = (v^(2))/(2g) = (( sqrt((10)/(7) gh))^(2))/(2g) = (10)/(14) h = (5)/(7) h`