Finding Distance Up (Max Height) trouble

[Mirth]

Hey guys, I just started a physics class in college and I'm having a hard time understanding most of it, but I'm trying...

Anyways, I'm trying to find "Distance Up" for the problem below, and I just can't understand how the professor got 1.28m for his answer.

Homework Statement

A projectile launched at an angle of 30 degrees, and a velocity of 10 m/s.
Find:

Horiz. Vel. (I got 8.66 m/s)
Vert. Vel. (I got 5 m/s)
Time Up (He gave us this, .51 sec)
Total Time (I got 1.02 sec)
Distance Up (Can't figure this out...)
Distance (I got 8.84 m)

Homework Equations

Distance Up = (Final Vel^2 - Initial Vel^2) / 2 * Acceleration
or = Initial Vel * Time + 1/2 * Acceleration * Time^2

The Attempt at a Solution

Well, I'm having a hard time figuring out how to get Final Velocity with this type of problem...

I mean, the initial velocity should 0, so when I plug that into the 2nd formula I get:
(0 * 1.02) + (1/2) * 9.8 * 1.02^2 = 5.09796

But the correct answer should be 1.28 m...

 

[neutrino]

Hi Mirth. Welcome to PF!

The Attempt at a Solution

I mean, the initial velocity should 0, so when I plug that into the 2nd formula I get:

No, it is not. It is given in the problem that the velocity of the projectile at launch is 100m/s. (This is the initial velocity.)

As for the final velocity, it's the same as the velocity of a ball (that you throw straight up) when it reaches the maximum height. (The ball you throw up is also an example of a projectile, but it doesn't travel horizontally like the object in the problem.)

 

[Mirth]

Hi Mirth. Welcome to PF!


No, it is not. It is given in the problem that the velocity of the projectile at launch is 100m/s. (This is the initial velocity.)

As for the final velocity, it's the same as the velocity of a ball (that you throw straight up) when it reaches the maximum height. (The ball you throw up is also an example of a projectile, but it doesn't travel horizontally like the object in the problem.)

Oh, I see I see.

I also messed up in my original post putting 100m/s for velocity, when it's supposed to be 10 m/s. I'll edit that now.

But anyways, so if the initial velocity is 10 m/s, then in the second formula, it should be:

10 * 1.02 + (1/2) * 9.8 * 1.02^2

And I get 15.29796 for an answer, so I'm not sure what I'm putting in wrong. :(

 

[neutrino]

I made a slight but significant error in my earlier post, so please forgive me. :)

The initial velocity in this case is the vertical component of the velocity with which it was thrown.

And for the time, you shouldn't be using 1.02s. Can you see why?

 

[Mirth]

I made a slight but significant error in my earlier post, so please forgive me. :)

The initial velocity in this case is the vertical component of the velocity with which it was thrown.

And for the time, you shouldn't be using 1.02s. Can you see why?

Hehe, it's no problem.

So you're saying Initial Velocity is the Vertical Velocity that I found, which is 5 m/s.

As for the time part... I guess I should be using the "Time Up" I got before (.51 sec) because the point of the highest distance is at the time it took to get up to that point, which is the halfway point...

If that is so though, then the formula would be: 5 * .51 + (1/2) * 9.8 * .51^2, which I get 3.82449, which is still wrong.

I'm really terrible at the thinking part of this. :(

 

[neutrino]

So you're saying Initial Velocity is the Vertical Velocity that I found, which is 5 m/s.

Yes.

As for the time part... I guess I should be using the "Time Up" I got before (.51 sec) because the point of the highest distance is at the time it took to get up to that point, which is the halfway point...

Correct again.

If that is so though, then the formula would be: 5 * .51 + (1/2) * 9.8 * .51^2, which I get 3.82449, which is still wrong.

What you have there is really a vector equation. (Have you studies those yet?) It takes into account the directions of certain quantities as well. As per convention, in an x-y coordinate system, the +y axis points "upwards." The initial velocity points up and is therefore positive, but what about gravity?

 

[Mirth]

We haven't really learned anything about this chapter yet in class, so I'm just trying to use my book as a source (it feels horribly unorganized...), so I guess we haven't studied vector equations yet.

As for your question, I guess gravity would be negative because it's is pulling the object down...

 

[neutrino]

I guess gravity would be negative because it's is pulling the object down...

And how would that affect the equation

Distance = Initial Vel * Time + 1/2 * Acceleration * Time^2

?

 

[Mirth]

And how would that affect the equation

It'd be a negative number; -9.8

?

That's the formula he gave us to find Distance for this problem. Even if I use the -9.8, I get a really odd number.

I think I'm just going to skip this problem.

 

[neutrino]

Really?

(0.5*0.51) - (0.5*9.8*(0.51)^2) = http://www.google.com/search?hl=en&...UG&q=5*0.51+-+(0.5*9.8*(0.51)^2)&btnG=Search"

 

[Mirth]

Really?

(0.5*0.51) - (0.5*9.8*(0.51)^2) = http://www.google.com/search?hl=en&...UG&q=5*0.51+-+(0.5*9.8*(0.51)^2)&btnG=Search"

Hrm, yeah, I've never seen that formula, and that's not what he gave us so that totally threw me off.

I'm going to have to look in the book some more for that. I appreciate your help!

 

[neutrino]

It's the same formula, except that I inserted -9.8, instead of +9.8.