# Angle, horizontal range, velocity

1. Apr 18, 2007

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!

Last edited: Apr 18, 2007
2. Apr 18, 2007

### tony873004

You are correct that the y-component of the velocity equals 10 m/s. sin(30)=0.5. 0.5*20=10.

Hint: instead of jumping straight to the formulas, just try to intuitively picture what is happening:

You know that gravity slows things down by 10 meters per second every second.

How long will it take something that is travelling 10 meters per second to be slowed to a stop if it is decelerating at 10 meters per second every second?

That gets you to the top. Double it to get the ball back to the ground. You need this answer before you can find the other parts.