Question

A body, projected with a certain kinetic energy,
has a horizontal range R. The kinetic energy will
be minimum at a position of the projectile when
its horizontal range is
(a) R.
(b) 3R4
(d) R14
(c) R2​

Answer 1

Maximum possible range: θ=45° in the range

Formula = R = $\frac{u^2Sin2@}{g}$

maximum height achieved: H = $\frac{u^2sin^245}{2g}$ = $\frac{u^2}{4g}$

potential energy at the highest point =  half of the KE at ground

So, K=$\frac{1}{2}$​mu²

So, KE at the top point is: K - $\frac{K}{2}$ = $\frac{K}{2}$

since KE is minimum at the maximum height, so, the horizontal distance is: also $\frac{R}{2}$