If an athlete takes jump with velocity u and at an angle θ with horizontal, then
\(H = \frac{u^2sin^2\theta}{2g}\)
and \(R=\frac{u^2\,sin\,2\theta}{g}\)
∴ \(\frac{R}{H}=\frac{u^2\,sin\,2\theta}{g}\times\frac{2g}{u^2sin^2\theta}\)
= \(\frac{2\,sin\,2\theta}{sin^2\,\theta}=\frac{2\times2\,sin\,\theta\,cos\,\theta}{sin^2\,\theta}\)
= 4 cot θ
or R = 4H cot θ
Thus, the span of the jump depends on the height of the jump and the angle at which the athlete jumps.