Projectile Motion - Particle Acceleration

[Osbourne_Cox]

1.Please note that in the following problem the particle has accelerations in both the x- and the y-directions. Therefore the x and y components of velocity keep changing. Initially (at time t = 0) a particle is moving vertically at 5.6 m/s and and horizontally at 0 m/s. The particle accelerates horizontally at 2.1 m/s2 . The acceleration of gravity is 9.8 m/s2 .

At what time will the particle be traveling at 58◦ with respect to the horizontal? Answer in units of s.

Homework Equations

The Attempt at a Solution

No clue!

 

[kuruman]

What are the relevant equations for accelerated motion in 2-dimensions?

 

[Osbourne_Cox]

Vxi=Vicostheta Vyi=Visintheta

Dx=Vxt

d=1/2(Vi+Vf)t

a=(Vf-Vi)/t

d=Vi(t)+1/2at^2

Vf^2=Vi^2+2ad


That's all I know

 

[kuruman]

Good. That's all you need. Look at the third equation. Solve it for vf in terms of the other quantities. Write two equations, one for the horizontal direction and one for the vertical direction

vxf = ...

vyf = ...

 

[Osbourne_Cox]

I'm sorry I don't understand

 

[Heur1]

Questions like this are ultimately about asking you to find a solution to a system of equations. The hard part is finding the equations to solve. The easy part is solving them.

Recall the techniques for solving a system of equations. You can substitute equalities (so, if a+c=3, and a/2=1, you could then restate the second equation as a=2, and substitute 2 in the first equation, allowing you to solve for c), add and subtract equations to eliminate variables, and so forth.

First, get clear on the problem. Draw a diagram. At the time you want, the vector of the particle will be at a 58 degree angle from the x-axis. What will the x and y components of that vector represent?

Remember that, if we're graphing velocity, tan θ = (velocity y) / (velocity x).

So the angle in the question is giving you some key information about the relationship between velocity-y and velocity-x at that time. Make that relationship as clear as possible. We know θ, so you can state the relationship much more clearly.

You want to find the value of t. How can we manipulate those equations, and what we know about the relationship of the velocity-y to velocity-x, to give us an equation in terms of t?

You know the values of θ, the initial velocity-x, the acceleration-x, the initial velocity-y, the acceleration-y. We do not know the final velocity of x or y, and we do not know displacement of the particle. We want to find the time.

 

[Osbourne_Cox]

I am still very confused.

 

[kuruman]

Third equation

a=(Vf-Vi)/t

Solve for vf in terms of the other quantities.

Is there anything in all of the above that you don't understand?

 

[Osbourne_Cox]

No, after reading it over a few more times I think I understand it better...but what I don't get is that the t value we are using is a zero? No other t is given...? Thats where I am confused.

 

[kuruman]

Set the time aside for a moment. When the particle is traveling at 58^o with respect to the horizontal, what relation is true between the velocity components?

 

[Osbourne_Cox]

I have no clue...I'm sorry, I'm terrible at this... :(

 

[kuruman]

Draw a right triangle for the velocity vector. The right sides are the x and y components, The angle the vector makes with respect to the x-axis is 58^o. What is a relation that has v_x, v_y and 58^o? If you are still clueless, you need to review and learn how to resolve a vector into its components.

 

[Osbourne_Cox]

I've seen these before:

Vxi=Vicostheta Vyi=Visintheta

This has all three, but I've never used it yet:

tan θ = (velocity y) / (velocity x).

 

[kuruman]

This has all three, but I've never used it yet:

tan θ = (velocity y) / (velocity x).

There is always a first. At the time you are looking for, the angle is 58^o and on the right side of this equation you have Vyf / Vxf, the ratio of the final velocity components. So if you can express Vxf and Vyf as a function of time (with "t" in it), then you can solve that equation for t. To do that start with the third equation written twice, one for the x direction and ine for the y direction.