Problem 3.62, Young and Freedman-On the Flying Trapeze

  • Thread starter james brug
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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?.
 

Answers and Replies

  • #2
Gib Z
Homework Helper
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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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