Projectile Motion: Solving for Initial Velocity and Height

[perfect_piccolo]

Having some trouble with the last part of this question:

A ball is projected horizontally from the edge of a table that is 1.11 m high, and it strikes the floor at a point 1.36 m from the base of the table. What is the initial speed of the ball?

How high is the ball above the floor when its velocity vector makes a 46.4o angle with the horizontal?


So I've got the answer to the first part (yea me) by doing this:

First I solved for time, and got an answer of 0.476 seconds (I know I'm cheating by not typing the wquation out but I'm not sure how to type out square roots and intiial velocities and all those lovely little signs)

Then I went:

Delta x = v(intitial)(cos theda int)t
v=1.36 m / 0.476 s
v = 2.86 m/s

So I've got my initial velocity of 2.86 m/s.


So my problem comes with this second part of the questionow high is the ball above the floor when its velocity vector makes a 46.4o angle with the horizontal?

So I think I would be looking for Vy at 46.4 degrees, but I'm not sure how to find that, because all of the equations I know only contain initial theda, not some random theda along the way. Is this solved witha projectile motion equation, is it a purely vector related question, or a combination of both? Should I be using other information in the question?

Any guidance is greatly appreciated!

Thanks!

 

[Doc Al]

Hint: What value of Vy will give you the required angle?

 

[perfect_piccolo]

hmmm I need more of a hint I'm afriad...I'm not sure how I can find Vy if I only have the angle...I don't know what the size of the vector is (do I?), or otherwise I could go
Vy=Vsin(theda)

 

[learningphysics]

hmmm I need more of a hint I'm afriad...I'm not sure how I can find Vy if I only have the angle...I don't know what the size of the vector is (do I?), or otherwise I could go
Vy=Vsin(theda)

If you know the angle is 46.4... what do you know about tan(46.4) ? or rather tan(-46.4)

 

[perfect_piccolo]

hmmm I know that tan(-46.4)=(Vy / Vx) So if I calculate that out I get tan -46.4 = -1.05, but I'm not sure what that means. Is that the Vy value, or do I still have steps to go?

 

[Doc Al]

You have another step to go. The tangent equals the ratio Vy/Vx. You found the tangent, and you know Vx, so you can solve for Vy.

 

[perfect_piccolo]

Ok this may sound really lame (don't worry, I'm used to it lol) but how do I know what Vx is? The only x value I know is the total change in distance along the x-axis, which is 1.36 m. That's not the Vx value is it? Or is the velocity that I found (2.86 m/s) Vx?

 

[Doc Al]

Or is the velocity that I found (2.86 m/s) Vx?

Yes. That's the horizontal component of the velocity, which remains constant (gravity affects only the vertical component).

 

[perfect_piccolo]

so if tan(-46.4)=(Vy / Vx) ;
-1.05 (2.86 m/s) = Vy
-3.00 m/s = Vy ?

 

[Doc Al]

Looks good to me.

 

[perfect_piccolo]

that's good :D But I'm still confused about how to answer the question.

How high is the ball above the floor when its velocity vector makes a 46.4o angle with the horizontal?

 

[Doc Al]

Answer this question instead: Where is the ball when its vertical speed is -3.00 m/s?

 

[perfect_piccolo]

I don't know how to answer that question :( Does it use a projectile motion equation?

 

[Doc Al]

Yes. You'll need the kinematic equation describing accelerated motion. The vertical component of the projectile's motion is accelerated.