Chapter 3, Problem 24P

(a)

To determine

The takeoff speed of an athlete.

Answer to Problem 24P

9.6 m/s

Explanation of Solution

Given:

The athlete make an angle of 28° with ground and horizontal range is 7.80 m.

Formula Used:

Horizontal Range :

  $R=\frac{u^2\sin 2\theta }{g}$

Calculation :

Angle is,

  $\theta \text{ =}\text{28°}$

Horizontal Range, R = 7.80 m

Substituting the given values

  $7.80=u^2\frac{\sin 2\left[28\right]}{10}$

  $\frac{7.80\times 10}{\sin 2\left[28\right]}=u^2$

  $u^2=94$

  $u=9.6\text{ m/s}$

Conclusion:

Thus, the takeoff speed of an athlete is 9.6 m/s.

(b)

To determine

The length of jump if speed is increased

Answer to Problem 24P

An athlete can jump to 8.4 m length.

Explanation of Solution

Given:

The speed is increased by 5%, athlete make an angle of 28°.

Formula Used :

Horizontal Range :

  $R=\frac{u^2\sin 2\theta }{g}$

Calculation :

Velocity is increased by 5%.

  $\frac{105}{100}\times 9.6=10.08\text{ m/s}$

Substituting the values,

  $R=10.08\frac{2\sin 2\left(28\right)}{10}$

  $R=8.4\text{ m}$

Conclusion :

Thus, the athlete can jump to 8.4 m longer if speed is increased by 5%.