- #1

- 14

- 0

## 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: