Problem:

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 0.600 m and the acceleration achieved from this position is 1.25 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.)

Solution:

We are required to solve for the distance in a standing broad jump. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:

• The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
• The jumper is on level ground.

The formula for the range is

\text{R}=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta _{\text{o}}}{\text{g}}

To find the initial velocity of the jump, vo, we shall use the kinematic formula from the crouch position to the time the person leaves the ground.

\text{v}_{\text{f}}^2=\text{v}_{\text{o}}^2+2\text{ax}

In this case, the final velocity will be the initial velocity of the jump.

\begin{align*}
 \text{v}_{\text{f}}=\sqrt{\left(0\:\text{m/s}\right)^2+2\left(1.25\times 9.81\:\text{m/s}^2\right)\left(0.600\:\text{m}\right)}=3.84\:\text{m/s}
\end{align*}

So, the initial velocity of the flight is 3.84 m/s. We can now use the formula for range.

\begin{align*}
\text{R}&=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta_{\text{o}}}{\text{g}} \\
\text{R}&=\frac{\left(3.84\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}\\
\text{R}&=1.50\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}