Chapter 6, Problem 89RE

Solution

To find the minimum distance between the ball and the person K.

  $17.6786\text{ feet}$

Given information:

The person S stands 80 ft to left of the bottom of a Ferris wheel of radius 60 feet. At that instance the person K is at 3’clock on the Ferris wheel. The person S throws a ball with a velocity of 100 ft/sec and an angle of with horizontal of ${70}^{\circ }$ .

Calculation:

The angle ${70}^{\circ }$ in radian equals $\frac{70}{180}\pi =\frac{7\pi }{18}$ .

Assume that the lowest point of the Ferris wheel is at the ground.

The parametric equation for the path of the ball is $x\left(t\right)=-80+100\cos \left(\frac{7}{18}\pi \right)t$ , $y\left(t\right)=-16t^2+100\sin \left(\frac{7}{18}\pi \right)t$ .

And, the parametric equation of the Ferris wheel is $x\left(t\right)=60\cos \left(\frac{\pi }{6}t\right)$ , $y\left(t\right)=60+60\sin \left(\frac{\pi }{6}t\right)$ .

Now, the minimum distance between the ball and the person K will be at a time when the ball is just above the head of the person. At that time $-80+100\cos \left(\frac{7}{18}\pi \right)t=60\cos \left(\frac{\pi }{6}t\right)$ .

The solution of $-80+100\cos \left(\frac{7}{18}\pi \right)t=60\cos \left(\frac{\pi }{6}t\right)$ is $t\approx \text{2}\text{.6545}$ .

Now, the minimum distance between the ball and the person is the difference in height of the ball and the person at time $t\approx \text{2}\text{.6545}$ sec.

Therefore, the minimum distance is $\left(-16{\left(\text{2}\text{.6545}\right)}^2+100\sin \left(\frac{7}{18}\pi \right)\left(\text{2}\text{.6545}\right)\right)-\left(60+60\sin \left(\frac{\pi }{6}\times \text{2}\text{.6545}\right)\right)\approx 17.6786\text{ feet}$ .

Conclusion:

The minimum distance between the ball and the person K is $17.6786\text{ feet}$ .