Sure!
Well, what you have to do is split the ball's motion up into two parts:
-It's vertical motion
-It's horizontal motion
Picture the ball's motion:
The hypotenuse of the triangle is the ball's full motion. The adjacent side is it's horizontal motion, and the opposite side is the vertical motion.
We know that the hypotenuse is 15 m/s. So the adjacent side is Sin(25) * 15 =
6.34 m/s, and the opposite side is Cos(25) * 15 =
13.59 m/s. These are the initial velocities.
The vertical motion is fairly easy to imagine, right? The ball will go up to a certain height, stop, then begin falling back down. The equation is the basic motion formula for constant acceleration:
x(t) = (1/2) at^2 + v0 t + x0
x0 is initial position, which is zero. (We're assuming the ground is position zero)
v0 is initial velocity, which is 6.33 m/s (we're talking about vertical motion only right now)
a is acceleration, which is -9.8 m/s^s (gravity)
t is time
so we have something that looks like:
x(t) = (1/2) * -9.8 t^2 + 6.33 t
Then we ask the question: at what time does the ball have a position of zero (on the ground)? Well there should be two answers, right? The ball starts off on the ground (t=0) and then it eventually lands once again. So let's solve that equation for when position ( x(t) ) is zero.
0 = (1/2) * -9.8 t^2 + 6.33 t
factor out a t
0 = (-4.9 t + 6.33 ) * t
t = 0 is a solution
t = 6.33 / 4.9 =
1.29 is a solution
So we know the ball landed 1.29 seconds after being kicked. We can use this to answer the questions, now.
a) How far does the ball travel (horizontally)? Well, it travels at a constant speed of 13.59 m/s, and keeps going for 1.29 seconds, so just multiply these together to get 17.53 m. (some error from my rounding)
b) How high does the ball go? Well, if the ball lands at 1.29 seconds, then it must reach it's highest point at half of that (since the motion is symmetrical). Thus the high point is at 1.29/2 = .645 seconds.
x(.645) = -4.9 (.645)^2 + 6.33 (.645) = 2.044 m (a little error b/c of my rounding)
Alternatively, if you don't want to assume that the motion is symmetrical, you could do this:
We know that when the ball is at it's high point, it isn't moving vertically. It momentarily comes to rest in the air. (Velocity = 0)
The equation for the velocity of the ball (vertically) is:
v(t) = a t + v0
plug in the values again...
v(t) = -9.8 t + 6.33
Solve for when v(t) = 0...
0 = -9.8 t + 6.33
t = 6.33/9.8 =
6.45 seconds, just like before. Solve for the position just like we did earlier, too.
Cool?
