Projectile Motion of a football

[smartins16]

Ugh! I have been working on this lab for hours now and I can not seem to make sense of my results. We went to the park and kicked a football, recorded its hang-time, and measured the distance it traveled. This is the information I have:

Time Change in x (m)
3.69 sec. 42.06

This is everything I know:

Change in x = 42.06 m Change in y = 0 m
a = 0 m/s a = -9.8 m/s^2(gravity)
Vx = ? Vy = ?
Time = 3.69 s Time: 3.69 s.

I need to find the initial(overall) velocity of the football as it is kicked. I need to find what this would be if it were kicked at a 30 degree angle, a 45 degree angle, and a 60 degree angle.

So far i have found that Vx = 11.4 because:
Change in X = VxT + 1/2aT^2
42.06 = Vx(3.69) + 1/2(0)(3.69^2)
Vx = 42.06/3.69
Vx = 11.4

I found Vy with the same formula and found it to be 18.081

I made a right triangle and gave it a 30 degree angle.


Sin30= 18.081/V (v being the overal velocity)
You get 36.162

This is where the problem hits. Technically, by taking the cos30 of 11.4, I should get the same number. But instead:
cos30=11.4/V
V = 11.4/cos30
V = 13.16

Please help me figure out what I'm doing wrong. I did examples in the book and everything worked fine with those numbers, I don't understand why they are not working with these. I know that the time and measurements are right bc it was given to us on the board.

Please help!

 

[ramcg1]

I need to find the initial(overall) velocity of the football as it is kicked. I need to find what this would be if it were kicked at a 30 degree angle, a 45 degree angle, and a 60 degree angle.
QUOTE]


For same horizontal distance or time of flight?

 

[wais]

Hi there,

I can understand your frustration with this lab. Projectile motion can be a tricky concept to grasp, but let's break down what you've done so far and see if we can find where the problem lies.

First, it's important to note that in projectile motion, the horizontal and vertical components of motion are independent of each other. This means that the horizontal distance traveled (change in x) and the vertical distance traveled (change in y) do not affect each other.

In your calculations for Vx and Vy, you used the formula d = Vt + 1/2at^2. This formula is used for calculating the distance traveled, but we need to use a different formula to find the initial velocity. The formula we need to use is V = V0 + at, where V0 is the initial velocity and a is the acceleration due to gravity (-9.8 m/s^2).

Let's start with finding Vx. The formula we need to use is Vx = V0x + axt. In this case, V0x is the initial horizontal velocity, which is what we are trying to find. We know that the acceleration in the horizontal direction is 0 m/s^2, so the formula becomes Vx = V0x + 0. This means that Vx is equal to the initial horizontal velocity, which we can calculate using the information given in the problem.

Vx = change in x / time
Vx = 42.06 m / 3.69 s
Vx = 11.4 m/s

Now let's move on to finding Vy. The formula we need to use is Vy = V0y + ayt. In this case, V0y is the initial vertical velocity, which we are trying to find. We know that the acceleration in the vertical direction is -9.8 m/s^2, so the formula becomes Vy = V0y - 9.8t. We have the value for Vy (18.081 m/s) and the time (3.69 s), so we can plug those in and solve for V0y.

18.081 m/s = V0y - 9.8 m/s^2 * 3.69 s
V0y = 18.081 m/s + 36.162 m/s
V0y = 54.243 m/s

Now, let's move on to the