How Do You Calculate the Correct Vertical Velocity Component of a Projectile?

[physicsnewby]

Homework Statement

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.

Homework Equations

X component: Vx cos theta
Y component: Vy sin theta

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?

 

[Doc Al]

Have you stated the problem completely and exactly as given?

 

[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.

 

[Doc Al]

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.)

 

[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?

 

[robphy]

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?

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.

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 its when the object hits the ground, then the Y component would be zero then, wouldn't it?

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

 

[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

 

[robphy]

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?

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?

 

[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.

 

[robphy]

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

 

[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

 

[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.)

 

[Doc Al]

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.

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

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

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?)

 

[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.

 

[physicsnewby]

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

 

[Feldoh]

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

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

 

[Doc Al]

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

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!

 

[Feldoh]

I think it would help if you had some equations...
http://en.wikipedia.org/wiki/Equations_of_motion#Linear_equations_of_motion

 

[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.

 

[robphy]

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 WHY?
a = 5.89m/s^2

Vf^2 = Vo^2 +2aX
Vf^2 = 65^2 +2(5.89) (540) WHY X?
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.

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].)

 

[Plastic Photon]

use conservation of energy with 125m set as a zero for potential energy

 

[Doc Al]

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

This is OK. You can use it to find the components of the final velocity.

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

Not needed.

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

Not sure what you're doing here. The acceleration is g downward, not g sin 37! (No inclined plane in this problem.)

Use kinematics to find the x and y components of the final velocity. Solve for each component separately. (Hint: Only one component is affected by gravity. Which one?)

 

[vij]

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

You have already found the horizontal component. The final vertical component of velocity just before hitting the ground can be found as usual since you know the initial vertical component (let us say, ‘u’, which is 39.1 m/s), the acceleration ‘g’ (which is 9.8 m/s^2) and the displacement (let us say, ‘s’, which is 125 m). Use the equation v^2 = u^2 + 2gs.
Remember, your origin is at the edge of the cliff. When you consider the motion from the moment the projectile reaches the edge of the cliff while returning, all the vectors u, g and s are directed downwards and are therefore taken as negative. You are getting the vertical component of velocity itself, which is asked for in the problem. Since you get the value of v^2, you have negative and positive roots. You have to accept the negative root since the final vertical component is downwards.
[If you would prefer to consider the entire motion from the moment the projectile is projected up, you can do so. Only thing is, you will have to take the sign of ‘u’ as positive and those of ‘g’ and ‘s’ as negative. Interestingly, you will get the same equation for v^2].