Question

An athlete performing a long jump leaves the ground at a$27.0^{\mathrm{o}}$angle and lands 7.80 m away. (a) What was the takeoff speed? (b) If this speed were increased by just 5.0%, how much longer would the jump be?

Short answer

(a) The take-off speed of the athlete is $9.72\mathrm{m}/\mathrm{s}$.

(b) On increasing the speed by 5%, the athlete jumps 0.79 m longer.

Step-by-step solution

Step 1. Definition of range of a projectile motion

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.

Step 2. Identification of the given data

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

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

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

Step 3. (a) Determination of the launch speed of the athlete

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" $R_1=\frac{u_0^2\sin 2\theta }{g}\dots \left(i\right)$

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

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

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

Step 4. (b) Determination of the new range of the athlete

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:

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

Step 5. (b) Determination of the difference in the two ranges

The difference in the two ranges of the projectile is:

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

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