# Simple Initial Velocity Problem

An Olympic long jumper leaves the ground at an angle of 18.3 ° and travels through the air for a horizontal distance of 8.86 m before landing. What is the takeoff speed of the jumper?

I'm completely dumbfounded on this one. What I have done is cut the distance in half to create a triangle and then solve all sides in order to find the y. Once I have the way I use the formula ( t = sq rt of 2y/g ) to get the time for the object to fall half way. I multiplied by two to get total time of flight (or what I thought was total time of flight.) Once I have the alleged total time of flight I divide 8.86 by it, giving me the velocity in the x direction. Using cos(18.3) = 8.09 m/s divided by the hypotenuse, or the initial velocity. Solving for the initial velocity left me with 8.52 m/s, which is wrong.

Can anyone offer any help? I can scan my work if needed, I know it's hard to read and think about math like this. Thanks in advance!

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PhanthomJay