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In figure, the angle of inclination of t...

In figure, the angle of inclination of the inclined plane is `30^@`. Find the horizontal velocity `V_0` so that the particle hits the inclined plane perpendicularly.
.

A

`V_0 = sqrt((2gH)/5)`

B

`V_0 = sqrt((2gH)/7)`

C

`V_0 = sqrt((gH)/5)`

D

`V_0 = sqrt((gH)/7)`

Text Solution

Verified by Experts

The correct Answer is:
A
a. In a direction along the inclined plane,
`0 = V_0 cos 30^@ - g sin 30^@ t rArr t = sqrt(3) V_0//g`
In a direction perpendicular to incline,
`- H cos 30^@ = -V sin 30^@t - 1.2g cos 30^@t^2`
Putting the value of t and solving, we get `v_0 = sqrt((2gh)/5)` .
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