# Question on Projectile Motion

1. Oct 1, 2005

### scorpa

Hello everyone,

I'm trying to do a question on projectile motion, and I think the first part I have done right, but after that it is going downhill for me. Here is the question and my work:

A shotputter releases the shot some distance above the level ground with a velocity of 12.0m/s @51 degrees above the horizontal. The shot hits the ground 2.08 seconds later. You can ignore air resistance.
(a) What are the components of the shot's acceleration while in flight?
(b) What are the components of the shot's velocity at the beginning and at the end of its trajectory?
(c) How far did she throw the shot horizontally?
(d) Why does the expression for R (R=(Vi^2sin(2)(initial angle))/g) not give the correct angle answer for part c?
(e) How high was the shot above the ground when she released it.

(a) For this part, you know that all of the acceleration once the shot leaves her hand is due to gravity, therefore her Ax=0 and her Ay=-9.81m/s^2

(b)For this part I just used V(x)=Vi(costheta)= (12.0cos51)=7.55m/s, and V(y)=Vi(sintheta)= 9.33m/s. At the beginning of the trajectory the velocities should be the same as at the end of the trajectory, expect the y component of velocity will at this point be negative, as it is travelling downwards.

(c) x=(vicostheta)t = (12.0cos51)2.08s=15.7m

(d) For this part I think that the above equation won't work because you are dealing with y values when you are wanting to find horizontal values. Not to mention the fact that you are starting at a y value above the origin (shot is released above the ground).

(e) This is where most of my troubles start. Even if my above answers are not correct, at least I could figure out a value for them. I have no idea how to do this one.

2. Oct 1, 2005

### zwtipp05

vy=0 when the projectile is at max height.

3. Oct 1, 2005

### scorpa

Yes that I did know, it's just that it says how high was the shot above the ground when she released it. how do you figure that out?

4. Oct 1, 2005

### Grogs

OK

You've made an assumption here that is not justified. Set up your kinematics equations and solve for vxf and vyf.

OK

They're giving you a hint about part (b) here. Look at the Range equation in your book (it should be the one listed above.) Look at where the equation comes from. It's based on a special case that differs from what you're given here. What is it?

Look at what you're given and look at the values you've computed so far in parts (a) and (c). You're looking for the height at which the shot was released. How is that represented in your kinematics equations? Do you have an equation that includes this value that you could use to solve for it?

5. Oct 1, 2005

### scorpa

They never really say where the range equation comes from, they just give it to you without explanation unfortunately.
I now realize why my assumption above was not justified, seeing as the projectile was fired at a distance above the ground. I will redo that part. Other than that I'm afraid I am still lost.

6. Oct 1, 2005

### Grogs

No problem. You've recognized your bad assumption. All textbooks aren't created equal and some have a bad habit of just giving you a formula without explaining where it comes from. It's only applicable in a special case when the initial and final height (yi and yf) are the same. Given this assumption, vi would equal vf, but that is incorrect here.

For 2-D kinematics, you should have 4 equations, one each for x, y, vx, and vy . You've already solved for xf in part c. You'll need the 2 expressions for velocity for part (b) and the equation for y in part (e).

If you haven't already done so, draw out the problem. Make sure you label the origin (x,y)=(0,0) for your coordinate system. Based on this information, you should be able to fill in all the variables and solve the equations.

7. Oct 1, 2005

### scorpa

So for part e all I need to do is use the equation for y? I understand that this will give me a y distance, but I'm afraid the part that I don't get is how it will show how far up from the ground the ball was thrown, I'm sorry this is probably such a stupid question but I really want to understand what I'm doing instead of just doing it because I was told it would work...haha.

8. Oct 1, 2005

### scorpa

Oh I just had an idea! If I found the y distance the ball would travel if it was launched from the ground, and the y distance it actually travels, and subtracted the two it would give the distance above ground from which it was launched! Does that make any sense? When i'm excited my typing tends to be incoherant ....haha.

9. Oct 1, 2005

### Grogs

I think you're almost there. Your explanation sounds a little convoluted though, but what you're describing is correct. Since you're so close, let's go ahead and look at the equation for y:

$$y_{f} = y_{i} + v_{ix}t + \frac{1}{2}gt^2$$

You were given g and t. You solved for vix in part (b). The only 2 unknowns at this point are yf and yi . If you set the origin for your coordinate system (x,y)=(0,0) at the point where the shot is released, what does that tell you about yi? Can you solve it now?

10. Oct 1, 2005

### scorpa

Well if you assumed the coordinate system was set at (0,0) wouldn't that tell you that yi is 0? I'm sorry but I don't see how that helps, maybe I'm just thinking about it the wrong way?

11. Oct 2, 2005

### Grogs

If yi is 0, you have only one unknown in the equation and that's yf . You can solve directly for yf . We know yf is on the ground, so if we let h be the distance above the ground the shot was released, h=yi - yf . For example, if you get -2.00m for yf (I just made this up), then h = 0 - -2.00m = 2.00m.

Remember, your choice of where to place your coordinate system is completely arbitrary. You could set y=0 on the ground, in which case yf would be zero and you could solve the equation for yi. You could even set your y=0 100m above the ground, in which case yf would be -100m. No matter how you set up your y-axis, h = yi - yf and you're going to get the same answer.

12. Oct 2, 2005

### scorpa

OK thank you I will give that a try! I really appreciate your help on this, you've been really helpful!

13. Oct 2, 2005

### scorpa

Ok, I tried this and I got my height above ground to be 5.52m. I don't know about you but this seems way to high!!!! I used the equation you gave me above, and that's the answer I got. What could have gone wrong?

14. Oct 2, 2005

### Grogs

Oops, you're right. I put vix in the equation when it should have been viy. It should look like this:

$$y_{f} = y_{i} + v_{iy}t + \frac{1}{2}gt^2$$

If you use vix you *will* get 5.52m, but of course that's not correct. You're obviously doing the math right though, so use viy and I think you'll get a slightly more realistic answer.

15. Oct 2, 2005

### scorpa

Haha!!! Ok, I was just going to ask why we were dealing with Vix!!!! That makes so much more sense now!!! Thank's again, you've been a great help!!!

16. Oct 2, 2005

### scorpa

Ah ok, 1.81m seems a little more realistic!!! Thanks!

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