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A small block slides with velocity 0.5sq...

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.

Text Solution

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The correct Answer is:
A, C, D
`v^(2)=v_(0)^(2)+2gh=(0.5sqrt(gr))^(2)+2gr(1-costheta)`
`=(2.25gr-2grcostheta)`
At the time of leaving contact, `N=0`
:. `mgcostheta=(mv^(2))/(r)=2.25mg-2mgcostheta`
:. `costheta=(2.25)/(3)=(3)/(4)`
:. `theta=cos^(-1)(3//4)`
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