A small block slides with velocity `0.5sqrt(gr)` on the horizontal frictionless surface as shown in the figure. The block leaves the surface at point `C` . Calculate angle `theta` in the figure.

Options

A. `cos^(-1).(4)/(9)`
B. `cos^(-1).(3)/(4)`
C. `cos^(-1).(1)/(4)`
D. `cos^(-1).(4)/(5)`

Answer

B

Explanation

At point `C mg cos theta - N = (mv^(2))/( r)`
(when block leavels the surface normal force becomes zero, so putting `N = 0) implies g cos theta = (v^(2))/( r)`
`v^(2) = v_(0)^(2) + 2 gh, h = r - r cos theta`
gr cos theta = (0.5 sqrtgr)^(2) + 2g (r - r cos theta)
`implies cos theta = (1)/(4) + 2 - 2 cos theta implies cos theta= (3)/(4)`