Problem 3.62, Young and Freedman-On the Flying Trapeze

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Mary Belle must achieve an initial speed of approximately 13.7 m/s to reach Joe Bob, who is positioned 6.1 m above and 8.2 m horizontally from her launch point. The calculations indicate her initial vertical velocity is 10.93 m/s, while her horizontal velocity is 8.24 m/s. However, the expected values from Mastering Physics are 14 m/s for initial speed, 11.2 m/s for vertical velocity, and 8.4 m/s for horizontal velocity. In the second part of the problem, it is confirmed that at the peak of her trajectory, her vertical velocity is indeed 0, aligning with the calculated horizontal component of 8.4 m/s. The discrepancies suggest possible errors in the textbook's 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 V_o, 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) V_y^2=V_{0y}^2+2a_y(y-y_0)

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

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

tan(53^{\circ}) =\frac{10.93}{x} \rightarrow x= 8.24 m/s 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, V_x. This means that Mary Joe is at the top of her parabolic trajectory, where her y-velocity, V_y 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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