# Homework Help: Components of Velocity

1. Jun 28, 2007

### physicsnewby

1. The problem statement, all variables and given/known data
I have to find the x and y components of velocity. I'm given the original velocity of 65 m/s and the projectile is shot at an angle of 37 degrees.

2. Relevant equations
X component: Vx cos theta
Y component: Vy sin theta

3. The attempt at a solution

X component: 65 cos 37 = 51.9 m/s
Y component: 65 sin 37 = 39.1 m/s

I get the right answer for cos, but the correct answer for sin is -63.1m/s. How do you get this answer?

2. Jun 28, 2007

### Staff: Mentor

Have you stated the problem completely and exactly as given?

3. Jun 28, 2007

### physicsnewby

This is the exact question:

A projectile is shot from the edge of a cliff 125m above ground level with an initial speed of 65m/s at an angle of 37 degrees. Determine the horizontal and vertical components of its velocity.

4. Jun 28, 2007

### Staff: Mentor

I assume they mean for you to determine the components of the velocity as it hits the ground. (Not the initial velocity, but the final velocity.)

5. Jun 28, 2007

### physicsnewby

Hmmm, I didn't think of that.

That said, I'm still not sure where the -63.1 is coming from. If its when the object hits the ground, then the Y component would be zero then, wouldn't it?

6. Jun 28, 2007

### robphy

Why are you using 37 degrees?
Did you draw a diagram of the trajectory?
(Even with an incorrect angle, do you know why you got the right numerical answer for Vx?)

If you were lying down at the impact point, what would you feel when that object arrives?

Last edited: Jun 28, 2007
7. Jun 28, 2007

### physicsnewby

I'm using 37 degrees because it was the angle in the question. There is a diagram in my text which I am also using for this.

What angle should I be using? I'm not sure what else to use.

Thanks

8. Jun 28, 2007

### robphy

On your diagram, draw in that 37 degree angle.
Now, your question asks [implicitly] for what happens at the impact. What is the angle there? Is it 37 degrees?

9. Jun 28, 2007

### physicsnewby

ah, good question!

At impact, I'm not sure what the angle is. From the picture it looks like its almost 90 degrees, but I'm pretty sure that can't be.

10. Jun 28, 2007

### robphy

So, you need a different way to determine the [components of the] final impact velocity. Any more "relevant equations" for this problem?

11. Jun 28, 2007

### physicsnewby

I am drawing a blank. I only thought you needed the init. velocity and then multiply by cos theta for one component and by sin theta for the other.

Using the final impact velocity and then doing cos and sin would make sense, but I don't know what angle to use.

I guess I somehow need to find the final velocity too

12. Jun 29, 2007

### robphy

Don't focus on finding that angle right now.
Can you find instead the magnitude of the final velocity? or possible easier, the components of the final velocity [without using the so-far undetermined angle of the final velocity]?

Note this is not a pure geometry problem. This is a physics problem. So, there are other physical principles involved. (Hint: there are more equations that are relevant to this problem.)

13. Jun 29, 2007

### Staff: Mentor

That would give you the components of the initial velocity, but you need the components of the final velocity.

Hint: Solve for the components of the final velocity. (You won't even need to figure out the angle.) Treat each component separately and apply the correct kinematic equations to solve for the final value. (Which way does gravity act?)

14. Jun 29, 2007

### vij

Write the X-component of velocity as Vcosθ and not as Vx cosθ. Similarly, write the Y-component as Vsinθ and not as Vysinθ.
Since you are asked to find the components of velocity, the question is simple. Gravity cannot change the horizontal component (X-component, as you have imagined). So, its value is 51.9 m/s ( as you have already calculated), through out the motion.
The vertical component (the Y-component, as you have imagined) will change because of the gravitational pull. When the projectile reaches the level of the cliff while returning from the top of its path, its vertical velocity component will have the same magnitude as that it had initially (39.1 m/s) when it was projected up. Why not use this as the initial vertical velocity and ‘g’ (equal to 9.8 m/s) as the acceleration and calculate the final vertical velocity?
The sign will be negative as the direction is downwards.

15. Jun 29, 2007

### physicsnewby

Once I find the final velocity, how does that help me find the component of velocity if I don't have an angle?

16. Jun 29, 2007

### Feldoh

They are telling you to find the components of the final velocity. If you know the components simple trig tells us that

$$\theta = tan^{-1}(\frac {v_{y}}{v_{x}})$$ For any point in the projectiles path of motion

Last edited: Jun 29, 2007
17. Jun 29, 2007

### Staff: Mentor

Note that the problem doesn't ask for the velocity or the angle--all they ask for is the velocity components, which you should be solving for directly. (Of course, once you have the components you can figure out the total velocity and the angle of impact, if you wanted to.)

Note the first step is finding the components of the initial velocity, which you've already done in your first post!

18. Jun 29, 2007

### Feldoh

19. Jun 29, 2007

### physicsnewby

not sure if this is right,

y = y0 + Vo sin t - 1/2gt^2
0 = 125 + 65 sin 37 - 4.9t^2
0 = 125 + 39.1 - 4.9t^2
through quadratic equation, t= 10.4s is time for projectile to hit ground

X= Xo +XVo t +1/2at^2
X = 0 + 65 cos 37 (10.4)
X = 540m (the distance projectile lands from base of cliff)

a = g sin 37
a = 5.89m/s^2

Vf^2 = Vo^2 +2aX
Vf^2 = 65^2 +2(5.89) (540)
Vf^2 = 102.3 m/s

I get final velocity of 102.3m/s, not sure what to do with this though. I'm still not 'getting' this.

20. Jun 29, 2007

### robphy

Go back and look at WHY?.
With your attempted method, once you find the impact speed, recall that the x-component of the velocity is constant. So, you can use the Pythagorean theorem to determine the y-component of the final velocity.

(You might have found it easier to use the component formulas for velocity
[gotten by taking the time-derivatives of your position equations].)