91Ó°ÊÓ

The range of a projectile motion or range of trajectory is the total displacement along the horizontal direction.The projectile does not experience any acceleration along the horizontal direction because gravity only acts vertically.

The range depends on the initial speed (launch speed) and the angle of projection.

The angle of projection is $\mathrm{θ}=27.0^{\mathrm{o}}$.

The range of the jump is $R=7.80\mathrm{m}$.

The acceleration is $a=g=9.8\mathrm{m}/{\mathrm{s}}^2$.

Consider the vertical motion of the athlete. Let the positive y-direction be upward.

Let the initial velocity of the athlete will be $u_0$.

The level horizontal range of any projectile is given by:

role="math" localid="1644396071260" $R_1=\frac{u_0^2\sin 2\mathrm{θ}}{g}…\left(i\right)$

From equation (i), the launch speed will be given by:

$\begin{array}{c}{u}_0=\sqrt{\frac{R_{1}g}{\sin 2\mathrm{θ}}}\\ =\sqrt{\frac{\left(7.80\mathrm{m}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{\sin \left({54.0}^{\mathrm{o}}\right)}}\\ =9.72\mathrm{m}/\mathrm{s}\end{array}$

Thus, the take-off speed of the athlete is $9.72\mathrm{m}/\mathrm{s}$.

In the second case, it is given that the launch speed is increased by 5%. Therefore, the new speed will be:

$\begin{array}{c}{u}_1=u_0+\frac{5}{100}{u}_0\\ =1.05u_0\\ =1.05\left(9.72\mathrm{m}/\mathrm{s}\right)\\ =10.2\mathrm{m}/\mathrm{s}\end{array}$

The new range of the athlete will be:

$\begin{array}{c}{R}_2=\frac{u_1^2\sin 2\mathrm{θ}}{g}\\ =\frac{{\left(10.2\mathrm{m}/\mathrm{s}\right)}^2\left(\sin {54}^{\mathrm{o}}\right)}{9.8\mathrm{m}/{\mathrm{s}}^2}\\ =8.59\mathrm{m}\end{array}$

The difference in the two ranges of the projectile is:

$\begin{array}{c}\mathrm{Δ}R=R_2-R_1\\ =8.59\mathrm{m}-7.80\mathrm{m}\\ =0.79\mathrm{m}\end{array}$

Thus, on increasing the speed by 5%, the athlete jumps 0.79 m longer.