91Ó°ÊÓ

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

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)

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

On substituting the given values in equation 1, we get

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

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 .$\mathit{θ}\mathbf{=}\mathbf{45}\mathbf{°}$

The equation for the range for a projectile is given by

$R=\frac{{V_i}^2\sin \left(2{\mathrm{θ}}_i\right)}{g}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦.(2)

Substituting the given values in equation 2, we get

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

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