Touching a hemispherical dome of radius `R` there is a vertical tower of height `H = 4 r`. A boy projects a ball horizontally at speed `u` from the top of the tower. The ball strikes the dome at a height `(R)/(2)` from ground and rebounds. After rebounding the ball retraces back its path into the hands of the boy. Then the value of `u` is :

Options

A. `sqrt(20gR)`
B. `sqrt(21gR)`
C. `sqrt(17gR)`
D. None of these

Answer

B

Explanation

The ball must strike the sphere normally (i.e. along radius) so that is can rebound back along the same path
`sintheta = (R//2)/(R) = (1)/(2)`
`theta = 30^(@)`
The velocity of the ball at the instant of hitting the sphere is inclined at `theta` to the horizontal
`:. tantheta = (v_(y))/(v_(x))`
`(1)/(sqrt(3)) = (sqrt(2g((7R)/(2))))/(u)`
`:. u = sqrt(21gR)`