Horizontal velocity?

Question

A broad jumper begins his jump at an angle of 22.0 degrees from the horizontal. At the middle of the jump he reaches a height of 70.0 cm. What is the horizontal velocity?

Answer 1

First, I need to find the vertical velocity. I will use this equation:

v^2=v(init)^2+2aD

Since the final velocity is going to be zero, we can drop the first part of the equation and get...
-v(init)^2=2aD

plugging the given values in, we get...
-v(init)^2=-13.734
v(init)=about 3.71 m/s

To find the horizontal velovity, we will use this equation:

tan22=3.71/v

solving for v, we get 9.1826 m/s as the horizontal velocity

Answer 2

I understand the sentence "At the middle of the jump he reaches a height of 70.0 cm" indicates the highest height is 70.0 cm.
Let his jump at an angle of 22.0 degrees is at velocity v. Hence his horizontal and vertical velocities are v*cos(22.0degree) and v*sin(22.0degree), respectively. Also, from energy conservation:
v*sin(22.0degree) = sqrt(2*g*h) = sqrt(2*9.8*0.700) = 3.70 m/s
Finally the horizontal velocity is:
v*cos(22.0degree) = 3.70 *{cos(22.0degree)/sin(22.0degree)}
= 1.50 (m/s)