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

If R is the horizontal range for theta inclination and h is the maximu

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