• Support PF! Buy your school textbooks, materials and every day products Here!

Initial Speed and Projectile Motion

  • #1

Homework Statement



The path of a golf ball over level ground reaches a height of 4.98m and a horizontal range of 24.8m. What is the ball's initial speed?
So it's a projectile motion problem. I draw a graph showing the parabolic trajectory of the ball, with the start point at the origin. It's final x = 24.8m and final y = 0m

Homework Equations



We have three constant acceleration equations we use in my course:

[tex]\vec{v}_{fx} = \vec{v}_{ix} + \vec{a}_{x}t[/tex]

Ensuring everything is in the same direction (vertical or horizontal), final velocity = initial velocity + acceleration * time

[tex]\vec{x}_{f} = \vec{x}_{i} + \vec{v}_{ix}t + \frac{1}{2}\vec{a}_{x}t^2[/tex]

Looks like an integral. Final position = initial position + initial velocity * time + half acceleration * t squared

[tex]\vec{v}_{fx}^2 = \vec{v}_{ix}^2 + 2\vec{a}_{x}Δ\vec{x}[/tex]

Final velocity squared = initial velocity squared + double acceleration * displacement

The Attempt at a Solution



My prof posted a video meant to give hints that will help us get started on the problem. I can try linking it here, though I'm not sure it'll work:

https://dal.echo360.com:8443/ess/echo/presentation/7b597979-18b5-413e-8a63-0aca1db8801d

Basically what it says is that, the motion is parabolic, I know the max height, and we know from class that the velocity at a projectile at max height is 0.

I then draw a graph with only half the parabola, starting at (x,y) = (0,4.98) and ending at (24.8, 0).

Then I have:

[tex]\vec{a}_{y} = -9.8m/s/s[/tex]
[tex]\vec{v}_{iy} = 0m/s[/tex]
[tex]\vec{y}_{i} = 4.98m[/tex]
[tex]\vec{x}_{f} = 24.8m[/tex]
[tex]t = ?[/tex]
[tex]vf = ?[/tex]

I use the second equation I listed above to find t:

[tex]0m = 4.98m + (0m/s)t + 1/2(-9.8m/s/s)t^2[/tex]
[tex]0m = 4.98m - (4.98m/s/s)t^2[/tex]
[tex](4.9m/s/s)t^2 = 4.98m[/tex]
[tex]t = 1.0081302s[/tex]

I keep in mind that this is only the time for half the parabola, so if I use this for the whole thing, I will need to double it.

We know from class also that the velocity at the end of a projectile's path is equal to but opposite in direction of the initial velocity. We also know that if we find the x and y components of the final velocity, we can add them to obtain the final velocity.

For the y component, I use equation 2 from above:

[tex]\vec{v}_{fy} = \vec{v}_{iy} + \vec{a}_{y}t[/tex]
[tex]\vec{v}_{fy} = 0m/s + (-9.8m/s/s)(1.0081302s)[/tex]
[tex]\vec{v}_{fy} = -9.87968m/s[/tex]

For the x component, I use equation 2 (acceleration along x is always 0m/s/s for projectiles, we learned in class):

[tex]\vec{x}_{f} = \vec{x}_{i} + \vec{v}_{ix}t + \frac{1}{2}\vec{a}_{x}t^2
[tex]24.8m = 0m + (\vec{v}_{ix})(1.0081302s) + 0[/tex]
[tex]24.8m = (1.0081302s)\vec{v}_{ix}[/tex]
[tex]\vec{v}_{ix} = 24.599997m/s[/tex]



This is where my main problem is. I'm guessing I did all of the above correctly, because it was simply subbing into the equations. I'm just not sure how to get the initial SPEED when I have the VELOCITY components.

Hopefully this is easy enough to understand, and thanks for all who read.
 
Last edited:

Answers and Replies

  • #3
Doc Al
Mentor
44,892
1,144
I then draw a graph with only half the parabola, starting at (x,y) = (0,4.98) and ending at (24.8, 0).
That first point is not part of the parabola. It reaches max height at half the horizontal range.


I use the second equation I listed above to find t:

[tex]0m = 4.98m + (0m/s)t + 1/2(-9.8m/s/s)t^2[/tex]
[tex]0m = 4.98m - (4.98m/s/s)t^2[/tex]
[tex](4.9m/s/s)t^2 = 4.98m[/tex]
[tex]t = 1.0081302s[/tex]

I keep in mind that this is only the time for half the parabola, so if I use this for the whole thing, I will need to double it.
OK.

For the y component, I use equation 2 from above:

[tex]\vec{v}_{fy} = \vec{v}_{iy} + \vec{a}_{y}t[/tex]
[tex]\vec{v}_{fy} = 0m/s + (-9.8m/s/s)(1.0081302s)[/tex]
[tex]\vec{v}_{fy} = -9.87968m/s[/tex]
OK.

For the x component, I use equation 2 (acceleration along x is always 0m/s/s for projectiles, we learned in class):

[tex]\vec{x}_{f} = \vec{x}_{i} + \vec{v}_{ix}t + \frac{1}{2}\vec{a}_{x}t^2
[tex]24.8m = 0m + (\vec{v}_{ix})(1.0081302s) + 0[/tex]
[tex]24.8m = (1.0081302s)\vec{v}_{ix}[/tex]
[tex]\vec{v}_{ix} = 24.599997m/s[/tex]
Don't forget that the time is for half the motion.



This is where my main problem is. I'm guessing I did all of the above correctly, because it was simply subbing into the equations. I'm just not sure how to get the initial SPEED when I have the VELOCITY components.
If you have the components of any vector, use the Pythagorean theorem to find the magnitude.
 
  • #4
Thanks to both of you. I don't know how I managed to miss that while reviewing my notes.

I also seem to have done the x component wrong. The initial x position was 12.4m, not 0m, which gave me the correct answer with the pythagorean theorem. Thanks again.

Now if I want the initial velocity of the ball, I can just add both velocity components, correct?
 
  • #5
Doc Al
Mentor
44,892
1,144
Now if I want the initial velocity of the ball, I can just add both velocity components, correct?
As long as you "add" them as vectors, sure.
 

Related Threads on Initial Speed and Projectile Motion

  • Last Post
Replies
12
Views
7K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
4K
  • Last Post
Replies
2
Views
2K
Replies
17
Views
5K
Replies
8
Views
5K
Replies
15
Views
2K
Replies
5
Views
4K
Top