In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is$\mathit{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{600}$m and the acceleration achieved from this position is$\mathbf{1}\mathbf{.}\mathbf{25}$times the acceleration due to gravity, g. How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

• Extension of the legs from the crouch position,$x=0.600$m,
• Acceleration,,$a=1.25g=1.25\times 9.8=12.25m/s^2$

where $g=9.8m/s^2$is the acceleration due to gravity.

We have the one-dimensional kinematic equation for the final velocity, v₂ ( or the velocity at which he leaves the ground, v_i)

${{\mathit{V}}_{\mathbf{2}}}^{\mathbf{2}}\mathbf{=}{{\mathit{V}}_{\mathbf{1}}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{a}\mathit{x}$…………………………………….(1)

On substituting the given values in equation 1, we get

$$\begin{gathered} {V_2}^2=0^2+2\times 12.25m/s^2\times 0.600m \\ =14.8m^2/s^2 \\ {V}_2=\sqrt{14.8} \\ =3.85m/s \end{gathered}$$

This velocity is equal to the velocity with which he is jumping from the ground. So

For a projectile motion, to reach a maximum distance (range), the angle with respect to the horizontal should be 45°, that is .$\mathit{\theta }\mathbf{=}\mathbf{45}\mathbf{°}$

The equation for the range for a projectile is given by

$R=\frac{{V_i}^2\sin \left(2{\theta }_i\right)}{g}$……………………….(2)

Substituting the given values in equation 2, we get

$$\begin{gathered} R_{max}=\frac{{(3.85)}^2\times \sin (2\times 45°)}{9.8} \\ =\frac{{(3.85)}^2\times \sin (2\times 45°)}{9.8} \\ =\frac{14.82\times \sin 90°}{9.8} \\ =1.51m \end{gathered}$$

The maximum horizontal distance covered or the distance up to which they can jump $R=1.51m$.