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A body is thrown with the velocity v0 at...

A body is thrown with the velocity `v_0` at an angle of `theta` to the horizontal. Determine `v_0 "in" m s^-1` if the maximum height attained by the body is `5 m` and at the highest point of its trajectory the radius of curvature is `r = 3 m`. Neglect air resistance. `[Use sqrt(80)= 9]`.

Text Solution

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The correct Answer is:
9
Radius of curvature at the highest point `r=v^(2)/g rArr r=((v_(0) cos theta)^(2))/(g) v_(0) cos theta=sqrt(gr)......(1)`
Maximum height, `h=(v_(0)^(2) sin^(2)theta)/(2g) rArr v_(0) sin theta=sqrt(2gh) .......(1)`
By using Eqs (i) and (ii), we get, `v_(0)^(2) sin^(2) theta+v_(0)^(2) cos^(2) theta =2gh+gr=q(2h+r)`
`rArr v_(0)=sqrt(g(2h+r))=sqrt(10(5+3))=9ms^(-1)`
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