Hi all, me with a few physics problems.

[BigGirl58]

Homework Statement

A dirt bike rider needs to complete a course that includes a "triple", a jump over three small hills that are speerated by a distance of 50 meters. The incline for the takeoff point is 37 degrees. How fast must the dirt bike rider travel to land 2.5 meters on the down side of the last hill?

Homework Equations

This is what I don't understand. I don't know what equation to use.

The Attempt at a Solution

None yet.

Homework Statement

A 75 kg paratrooper blacks out when experiencing downward forces in excess of 1000N. If he jumps out of a plane at 4000 meters and deploys his parachute at 2500 meters and the parachute causes an upward acceleration of 6 meters per second^2, will he black out? Why or why not?

Homework Equations

This is what I don't understand. I don't know what equation to use.

The Attempt at a Solution

None yet.

Homework Statement

A mobile TV camera is strung over an NFL stadium. The camera has a mass of 100 KG. There's a glitch in the rigging so the camera stalls and hangs down so that the angle between the 2 supporting cables is 45 degrees on the horizontal. What is the force (tension) on each cable?

Homework Equations

This is what I don't understand. I don't know what equation to use.

The Attempt at a Solution

None yet.


I've solved most of my problems. But these ones I don't understand how to approach. I'm not specifically looking for answers but just the proper way to attempt them. Answers are nice to check my eventual answers but I'll be attempting them myself either way. Thank you very much for reading!

 

[Hootenanny]

Lets take question one first, do you know any kinematic equations?

 

[BigGirl58]

Lets take question one first, do you know any kinematic equations?

Hmmm sounds familiar. That has to do with free-falling right?

 

[qspeechc]

Equations of motion. Do you know them?

 

[BigGirl58]

Equations of motion. Do you know them?

Somewhat I guess. Am I asking questiosn that are really basic? I'm very new to Physics is all. So many equations it's hard to figure out which goes where.

 

[Hootenanny]

Somewhat I guess. Am I asking questiosn that are really basic? I'm very new to Physics is all. So many equations it's hard to figure out which goes where.

Which kinematic equations [sometimes referred to as SUVAT equations] do you know?

 

[BigGirl58]

Which kinematic equations [sometimes referred to as SUVAT equations] do you know?

I only have these ones in my notes:

Vfinal^2 = Vinitial^2 + 2 * a * d

Vfinal = Vinitial + a * t

 

[Hootenanny]

I only have these ones in my notes:

Vfinal^2 = Vinitial^2 + 2 * a * d

Vfinal = Vinitial + a * t

Not have any relating S, u, a and t?

 

[BigGirl58]

Not have any relating S, u, a and t?

Unfortunately, no. I'm a great notetaker and there's not an S or U anywhere here.

 

[Hootenanny]

Okay, nevermind, so the first step would be to split the take off velocity into vertical and horizontal components.

 

[BigGirl58]

Okay, nevermind, so the first step would be to split the take off velocity into vertical and horizontal components.

Hmm...so you mean how fast and how high?

 

[Hootenanny]

Hmm...so you mean how fast and how high?

Nope, I mean splitting the velocity into a vertical velocity and horizontal velocity, we can write the vertical velocity thus;

[tex]v_y = v\sin\theta[/tex]

Do you follow?

 

[BigGirl58]

Nope, I mean splitting the velocity into a vertical velocity and horizontal velocity, we can write the vertical velocity thus;

[tex]v_y = v\sin\theta[/tex]

Do you follow?

Ohhh okay yea, I follow. Sorry I'm blonde.

 

[Hootenanny]

Ohhh okay yea, I follow. Sorry I'm blonde.

So am I, although I'm not a women...

So, what would be the horizontal component of velocity?

 

[BigGirl58]

So am I, although I'm not a women...

So, what would be the horizontal component of velocity?

would it be [tex]h_y = h\cos\theta[/tex]

 

[Hootenanny]

would it be [tex]h_y = h\cos\theta[/tex]

Looks okay to me. Now, using the vertical velocity can you work out how long he would be in the air for?

 

[BigGirl58]

Looks okay to me. Now, using the vertical velocity can you work out how long he would be in the air for?

Well, I know the Sin of 37 is .6 and the Cos is .8, but what is H and V?

 

[Hootenanny]

Well, I know the Sin of 37 is .6 and the Cos is .8, but what is H and V?

Lets stick you v's ok so;

[tex]v_x = v\cos\theta[/tex]

where v is the magnitude of the velocity.

 

[BigGirl58]

Lets stick you v's ok so;

[tex]v_x = v\cos\theta[/tex]

where v is the magnitude of the velocity.

I'm sorry, I don't get it. Is V an actual number I'm missing? Thank you for being so patient.

 

[Hootenanny]

Yes, v is the number you're missing.

And no problem

 

[BigGirl58]

Yes, v is the number you're missing.

And no problem

I guess I'm confused. The problem asks what speed the rider needs to go so how do we determine velocity if we don't know how fast he'll be going?

 

[Hootenanny]

I guess I'm confused. The problem asks what speed the rider needs to go so how do we determine velocity if we don't know how fast he'll be going?

Yes, but we know the distance he needs to travel...

 

[BigGirl58]

Yes, but we know the distance he needs to travel...

Is it 102.5 meters? Two 50m gaps and the 2.5 at the end?

 

[Hootenanny]

I would say you were almost right, I would say the horizontal distance is;

[tex]d = 2\times 50 + 2.5\cos\theta[/tex]

Since we're landing on a downward slope. Do you follow?

 

[BigGirl58]

I would say you were almost right, I would say the horizontal distance is;

[tex]d = 2\times 50 + 2.5\cos\theta[/tex]

Since we're landing on a downward slope. Do you follow?

Yea. So d = 102. I still don't see how we get that v though.

 

[Hootenanny]

Yea. So d = 102. I still don't see how we get that v though.

Well we know that since we have no acceleration in the horizontal direction;

[tex]v_x = \frac{d}{t}[/tex]

Now, let's look at the vertical direction using your kinematic equations. How would you work out the time of flight?

 

[BigGirl58]

Well we know that since we have no acceleration in the horizontal direction;

[tex]v_x = \frac{d}{t}[/tex]

Now, let's look at the vertical direction using your kinematic equations. How would you work out the time of flight?

ahh, I don't know. I don't get it. I don't understand how I can do that without knowing the velocity. What is that t?

 

[Hootenanny]

ahh, I don't know. I don't get it. I don't understand how I can do that without knowing the velocity. What is that t?

t is for Time.

Okay, forget about the horizontal direction, just consider the vertical direction; we do we not about the vertical velocity when the rider is at his maximum height?

 

[BigGirl58]

t is for Time.

Okay, forget about the horizontal direction, just consider the vertical direction; we do we not about the vertical velocity when the rider is at his maximum height?

That it will be 0.

 

[Hootenanny]

That it will be 0.

Correct, so using v_f = v_i+at we can write;

[tex]v -gt = 0 \Rightarrow t = -\frac{v}{g}[/tex]

Do you follow?

 

[BigGirl58]

Correct, so using v_f = v_i+at we can write;

[tex]v -gt = 0 \Rightarrow t = -\frac{v}{g}[/tex]

Do you follow?

okay yea, I follow.

 

[Hootenanny]

So, when the rider lands what will his vertical velocity be?

 

[BigGirl58]

So, when the rider lands what will his vertical velocity be?

I'm sorry, I don't know. It seems like we're going in circles. I'm looking at this stuff and d is the only number I have. Everything else is equal to other thigns I don't know. Like T, I understand that t = -v/g, but how do I get v? is g just 9.8?

 

[BigGirl58]

Anyone else care to help on a different problem?