EBK PRECALCULUS:GRAPHICAL,...-NASTA ED.
10th Edition
ISBN: 9780135795910
Author: Demana
Publisher: PEARSON CO
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Expert Solution & Answer
Chapter 6, Problem 89RE
Solution
To find the minimum distance between the ball and the person K.
17.6786 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∘ .
Calculation:
The angle 70∘ in radian equals 70180π=7π18 .
Assume that the lowest point of the Ferris wheel is at the ground.
The parametric equation for the path of the ball is x(t)=−80+100cos(718π)t , y(t)=−16t2+100sin(718π)t .
And, the parametric equation of the Ferris wheel is x(t)=60cos(π6t) , y(t)=60+60sin(π6t) .
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+100cos(718π)t=60cos(π6t) .
The solution of −80+100cos(718π)t=60cos(π6t) is t≈2.6545 .
Now, the minimum distance between the ball and the person is the difference in height of the ball and the person at time t≈2.6545 sec.
Therefore, the minimum distance is (−16(2.6545)2+100sin(718π)(2.6545))−(60+60sin(π6×2.6545))≈17.6786 feet .
Conclusion:
The minimum distance between the ball and the person K is 17.6786 feet .
Chapter 6 Solutions
EBK PRECALCULUS:GRAPHICAL,...-NASTA ED.
Ch. 6.1 - Prob. 1QRCh. 6.1 - Prob. 2QRCh. 6.1 - Prob. 3QRCh. 6.1 - Prob. 4QRCh. 6.1 - Prob. 5QRCh. 6.1 - Prob. 6QRCh. 6.1 - Prob. 7QRCh. 6.1 - Prob. 8QRCh. 6.1 - Prob. 9QRCh. 6.1 - Prob. 10QR
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