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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