Problem 3.62, Young and Freedman-On the Flying Trapeze

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SUMMARY

The discussion revolves around Problem 3.62 from Young and Freedman's physics textbook, focusing on the Texas Tumblers trapeze act. The initial speed required for Mary Belle to reach Joe Bob, positioned 6.1m vertically and 8.2m horizontally from her launch point, is calculated to be approximately 13.7 m/s, although Mastering Physics suggests values of 14 m/s for the initial speed and 8.4 m/s for her x-velocity component at the peak of her trajectory. The calculations involve using kinematic equations to determine the necessary initial velocities in both the x and y directions. The consensus is that the textbook may contain inaccuracies in the provided values.

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Problem 3.62, Young and Freedman--On the Flying Trapeze

A new circus act is called the Texas Tumblers. Mary Belle swings from a trapeze, projects herself at an angle of 53 degrees, and is supposed to be caught by Joe Bob whose hands are 6.1m above and 8.2 horizontally from her launch point. You can ignore air resistance.

(a) What initial speed [tex]V_o[/tex], must Mary Belle have just to reach Joe Bob?

(b) For the initial speed calculated in part a, what are the magnitudes and direction of her velocity when Mary reaches Joe?


My attempt at a solution

(a) [tex]V_y^2=V_{0y}^2+2a_y(y-y_0)[/tex]

[tex]-V_{0y}^2=-19.6m/s^2(6.1m)[/tex] set [tex]V_y[/tex] equal to 0

[tex]V_{0y}=10.93m/s[/tex] initial y velocity

[tex]tan(53^{\circ}) =\frac{10.93}{x} \rightarrow x= 8.24 m/s[/tex] x velocity

magnitude=13.7 m/s

Mastering Physics says these values should be 14m/s,11.2 m/s, and 8.4m/s.


(b) The magnitude according to Mastering Physics is 8.4m/s, which is also her x-velocity component, [tex]V_x[/tex]. This means that Mary Joe is at the top of her parabolic trajectory, where her y-velocity, [tex]V_y[/tex] is 0, right?.
 
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Well, I got the same values as you did, and to your (b), yes. I guess the textbook just got it wrong.
 

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