Question

In long jump, is the importance of this factor that how height you take before jumping?

Answer 1

Range \(R = \frac{u^2\sin 2\theta}{g}\)

For maximum range sin 2θ = 1 or 2θ = 90°

⇒ θ = 45°

For this angle of projection, maximum height

\(H = \frac{u^2\sin^245^\circ}{2g} = \frac{u^2}{4g} = \text{constant}\) 

Thus to attain θ = 45°, \(H = \frac{u^2}{4g}\) is necessary

This is why, in long jump, the player takes some height before jumping.