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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.
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Text Solution

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The correct Answer is:
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`0=v_(0)cos 30^(@)-g sin 30^(@) t rArr(v_(0)cos 30^(@))/(g sin 30^(@))`...(1)
`-H cos 30^(@)=-v_(0) sin 30^(@)t-1/2g cos 30^(@) t^(2)`...(2)
By equation (1) and (2), we get
`H=v_(0)^(2)/g[1+(cot^(2)alpha)/2]rArrv_(0)=sqrt((2gH)/5)=4m//s (alpha=30^(@))`
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