91Ó°ÊÓ

Powell’s jump is 8.95m

Powell’s speed at the time of takeoff is $v_0=9.50\mathrm{m}/\mathrm{s}$

Powell’s gravitational acceleration$g=9.8\mathrm{m}/{\mathrm{s}}^2$

This problem deals with the horizontal range of the particle which is nothing but the horizontal distance from the launch point at which the particle returns to the launch height. Using the formula for the range of the particle, the maximum range of the particle can be found. It is possible to determine how much less was Powell’s range than the maximum possible range for a particle, when they are launched at the same speed.

Formula:

The particle’s horizontal range can be given as,

$R=\frac{v_0^2}{g}\sin 2∆$ (i)

Where $v_0$ is the initial velocity of the particle.

$R=R_{max}when∆=45°$

Using equation (i), the maximum range will be,

$$\begin{gathered} R_{\max }=\frac{9.50\mathrm{m}/\mathrm{s}}{9.8\mathrm{m}/{\mathrm{s}}^2}\sin \left(90°\right) \\ {R}_{max}=9.209\mathrm{m} \end{gathered}$$

It is given that,

$$\begin{gathered} ∆R=R_{max}-8.95\mathrm{m} \\ ∆\mathrm{R}=0.259\mathrm{m} \end{gathered}$$

Therefore, Powell’s range was 0.259 m was less than the maximum possible range for a particle launched at the same speed.