What is the Takeoff Angle for a Long Jumper?

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SUMMARY

The discussion focuses on calculating the takeoff angle for a long jumper who rises 0.5 meters during the flight phase with a forward velocity of 8 m/s. The initial approach incorrectly interprets the 8 m/s as the hypotenuse of a right triangle, while it should be considered the horizontal component of the velocity. To accurately determine the takeoff angle, one must recognize that the 0.5 m is a vertical distance, not a velocity, and use the appropriate trigonometric relationships to solve for the angle, which is approximately 21.6 degrees.

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"A long jumper rises 0.5 meters during the flight phase of his jump. His forward velocity is 8 ms. What is the angle of take off?"

Okay, I did a skectch to work out basic trig, with 8m/s as my H and 0.5m as my 0
using sinϑ=o/h, =0.5/8 = 3
but it should be 21.6 ... do you know where I went wrong??
 
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I see a couple of problems so far. The first is that I would interpret the 8 m/s as the horizontal component of the velocity. Therefore if you draw a right triangle consisting of the velocity vector and its two components, the 8 m/s is not the hypotenuse of the triangle, but rather the horizontal leg.

The second problem is that you don't know what the vertical component is. You have to be careful here -- the 0.5 m provided is a distance, not a velocity. It's a little ambiguous what the "flight phase" of the jump is, but if you assume 0.5 m to be the maximum height that he reached, then you have enough info to solve the problem.
 

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