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What is the angle of projection to have a maximum range in 'Kitti pull'? If one strikes kitti pull with the speed of `98ms^(-1)`. What is the maximum range achieved?

Text Solution

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For maximum range, angle of projection `theta=45^(@)`.
maximum range `R_("max") = u^(2)sin2theta/(g)`
`R_("max")=(98xx98)/(9.8)=980m`
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