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If R is the horizontal range for theta ...

If R is the horizontal range for `theta` inclination and h is the maximum height reached by the projectile, show that the maximum ragne is given by `(R^(2))/(8h) + 2h`.

Text Solution

Verified by Experts

We know that horizontal range
`R=(u^(2)sin 2 theta)/g` and maximum height `h=(u^(2)sin^(2)theta)/(2g)`
`:.(R^(2))/(8h)+2h=([(u^(2)sin 2 theta)/g]^(2))/(8[(u^(2)sin^(2)theta)/(2g)])+2[(u^(2)sin^(2)theta)/(2g)]`
`=(u^(4)(2 sin theta cos theta)^(2))/(g^(2)xx8(u^(2)sin^(2)theta)/(2g))+(u^(2)sin^(2)theta)/g`
`=(u^(2))/g(cos^(2)theta+sin^(2)theta)=(u^(2))/g=R_("max")`
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