1. The problem statement, all variables and given/known data A soccer player kicked a ball up at a 30-degree angle at 20 m/s. Find the hang time, max. height and horizontal range. 2. Relevant equations x(sub y)=(1/2)at^2 + v(sub i sub y) Meaning... displacement = half of acceleration times time squared plus initial velocity in the y-direction. v(sub f)^2 = v(sub i)^2 + 2ax delta X = v(sub x) * T (sub y) 3. The attempt at a solution For the first equation, x(sub y)=(1/2)at^2 + v(sub i sub y) displacement is 0 because horizontally, it goes from the ground and ends back at the ground; no change. half of acceleration of -9.8 m/2^2 is 4.9 m/s^2. Now for the initial velocity, it says that it's in the y direction.. so you're supposed to set up a triangle from the curved path of the ball with the angle of 30 degrees, x on the bottom, y on the opposite of the 30d, with the hypotenuse being 20 m/s. When you take the sin 30 = y/20, you get y=10, so velocity in the y direction is ten. First of all, this was unclear to me because if you draw the triangle, it looks like y should be the maximum height. Anyway, so you plug those numbers into the equation, giving you. 0=-4.9(m/s^2) * t^2 + 10 (m/s) -10=-4.9t^2 2.04082 = t^2 1.43=t But the answer is supposed to be two seconds. Am I missing something here? Then after that, you want to find the max. height so you take v(sub f)^2 = v(sub i)^2 + 2ax and plug in numbers. 0 = 10^2(m/s) + 2(-9.8 m/s^2) * x 0 = 100 m/s - 19.6x -100 m/s = -19.6x 5.10 m = x Now comes the horizontal range, when you use delta X = v(sub x) * T(sub y) To find velocity in the x direction, you go to the triangle mentioned above and use cosine of 30 degrees instead, ending up with x=17.3 m/s. So delta x = 17.3 m/s * 2 s, which gives you delta x=34.6 m but everyone else got 35.35. Again, am I missing something? Thanks for the help!