Posted by penkey on Thursday, July 31, 2008 at 7:38pm.
An Olympic long jumper leaves the ground at an angle of 23 degrees and travels through the air for a horizontal distance of 8.7 meters before landing. What is the takeoff speed of the jumper?

physics  GK, Thursday, July 31, 2008 at 9:27pm
Let the initial velocity = Vo
The horizontal comonent, Vx = Vo(cos23)
The horizontal distance, x = Vo(cos23)t = 8.7m
t = 8.7m / Vo(cos23)
The vertical component of initial velocity is Vyo:
Vyo = Vo(sin23)
Let the final vertical velocity be Vyf. Then,
Vyf = 0 (a the highest position)
0 = Vo(cos23)  9.8m/s^2*t, or:
t = Vo(cos23) / 9.8m/s^2
You have two expressions for time, t.
Since the two are equal, you can solve for Vo

physics  W, Monday, October 25, 2010 at 4:52pm
Well this must be a pretty common question because I had it too. The problem here is the Voy is Vo(SIN(23) not cos)/10.0 (or 9.8).
Also because the Vfy is 0 in the middle of the jump used only use half the X for the horizontal component. This is how I got the 11 m/s initial take off speed.
I don't know for sure, but that's how I'm solving these type of questions.
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