What is the Takeoff Angle for a Long Jumper?

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The discussion focuses on calculating the takeoff angle for a long jumper who rises 0.5 meters with a forward velocity of 8 m/s. A participant initially misinterprets the velocity as the hypotenuse of a right triangle, but it should be considered the horizontal component. The vertical component of velocity is unknown, and the height of 0.5 meters represents maximum height, not velocity. To find the correct angle, one must use the appropriate trigonometric relationships involving both horizontal and vertical components. Clarifying these points is essential for accurately determining the takeoff angle.
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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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