Done prob, mult times and can't find the correct answer. help !

1. Oct 2, 2007

So I've been working on this problem for ALMOST a week now and I can't get the right answers for parts A and C. Although for some reason by answer for B is correct, which should make part A correct but it doesn't. I basically need a very helpful person to just do the problem for me, since no matter what I do I can't get the right answer. I'm using all the correct equations, and I've even had someone redo it to make sure I was using the right information, but it still doesn't work. I put the correct answer in part B, so hopefully I can get down tot he bottom of this. I'd really appreciate it, you have no idea! Thanks!

An airplane, diving at an angle of 52.0° with the vertical, releases a projectile at an altitude of 550 m. The projectile hits the ground 6.00 s after being released. (Assume a coordinate system in which the airplane is moving in the positive horizontal direction, and in the negative vertical direction. Neglect air resistance.) (PART B IS CORRECT, I've tried this multiple times and parts A and C won't come out correctly for anything!)
(a) What is the speed of the aircraft?
_______ m/s

(b) How far did the projectile travel horizontally during its flight?

(c) What were the horizontal and vertical components of its velocity just before striking the ground?
______ m/s (horizontal)
______ m/s (vertical)

2. Oct 2, 2007

dynamicsolo

Part of the reason the forum guidelines ask the student to show their work is that it makes it easier for the reader to spot possible errors if they know what the student tried to do. It's hard for us to tell you what you may have done wrong if we can't see what you did... (For example, not two hours ago, I helped a student with a calculation that didn't work because they had their calculator set in the wrong mode for entering angles.)

What do your calculations look like for all three parts? Thanks!

(Besides, I'm curious to see how you got part B to come out right without part A being right. Without seeing what you did, I'd say to be careful about how you used that 52º angle in your calculations.)

[later edit...]

I also get nearly this result (I used g = 9.81 m/(sec^2)), which makes it all the more puzzling why you and your colleague don't get part A to work. The plane and then the projectile are moving at an angle 52º to the vertical or 38º to the horizontal (if I'm reading the problem statement correctly). If you're getting part B right, you would now know the horizontal velocity component of the projectile's and the plane's speed. How could you use that to find the magnitude of the plane's speed?

If you solved this a completely different way from what I used, you may not follow what I'm getting at in that last question, which is why I'd like to see what you did...

Last edited: Oct 2, 2007
3. Oct 3, 2007

Sorry about that, but I'll include my work below. But yes I don't undestand how part B is right without part A being correct. In order to get B the A component must be right, but the program marked me as incorrect for that.

So for part a I used V-Vo=(VoSin)t-(.5)(g)(t^2) and ended up with Vo=79.0175. I did the equation previous times before and got different answers too, but this one helped me get the correct distance for part B.

But for some reason in part b I used the equation: Xf=VxiT and got.. 79.0175(6 seconds)= 474.105 m which is right.

Now for part C, finding the horizontal and vertical components of its velocity just before striking the ground were wrong too.
Vx=79.0175(sin52)=62.27
Vy=79.0175(cos52)=48.65

And yet... I'm still at a loss. Hope this gives you a better understanding of the problem and what I may be doing wrong.

4. Oct 3, 2007

dynamicsolo

Ah! That's why we ask you to show us what you did. Part (b) is correct because it uses the horizontal component of the plane's and the projectile's velocity; that is not, however, the complete velocity. (In fact, you changed what you called it between parts (a) and (b).)

Part of the problem is that you didn't use the initial altitude of the projectile for anything. I'm still sort of puzzled, in fact, how you manage to get anything useful in part (a); is there some other calculation that isn't posted here? You want to write the kinematic equation for the vertical position of the projectile to start; we'll go on from there...

5. Oct 4, 2007

hotvette

Here is what I think is going on:

1. $g = 9.8 \ m/s^2$ was used which is fine

2. There's a typo in the equation used for a. It should be $y-y_0=v_0\sin(\theta)t-0.5gt^2$ but appears to have been used correctly in the calculation, meaning the initial height was taken into consideration.

3. If $\theta=52^o$ is used, then $v_0 = 79.0175$. But, I believe the angle is wrong (I agree with dynamicsolo).

4. For b, the equation $x=v_xt$ was used, which is correct, but $v_0$ from 3 was used instead of $v_x$ which results in almost the correct answer for b. Net is, wrong value for $v_x$ gives nearly the correct value for b.

What puzzles me is why 474.105 is considered correct for b. I get 101.14 for a and 478.19 for b.

Last edited: Oct 4, 2007
6. Oct 4, 2007

dynamicsolo

That's why I was waiting to see the calculations. I also got 101.1 m/sec for part (a) and 478.0 m for part (b). My suspicion is that since 52º is near 45º, sin 52º and cos 52º = sin 38º got exchanged, but manage to give similar results.

(checks equations again) Yes, if you use $y-y_0=v_0\sin(\theta)t-0.5gt^2$ with y-y0 = 550 m, you can solve to get v0·sin(theta) = 62.24 m/sec, which is v_y. But if the plane is flying at an angle 52º off the vertical, then this would be

v_y = v_0·cos 52º = v_0·sin 38º , which tells us that

v_0 = 62.24 / 0.6157 = 101.1 m/sec.

Using sin 52º instead leads to v_0 = 78.98 m/sec , so this is the source of the controversy (or error) depending on whether we are understanding the description in the problem correctly.

For our result, v_x = v_0 · sin 52º = 79.67 m/sec
and an answer for part (b) of x = (79.67 m/sec)(6 sec) = 478.0 m.

Blink691's result is v_y = v_0 · sin 52º = 62.24 m/sec ,
so v_0 = 78.98 m/sec ,
but this then got used as v_x to get (78.98 m/sec)(6 sec) = 473.9 m .
This last step, I'm afraid, just isn't right. Things got very tangled here: it wasn't only that angles got swapped...

7. Oct 5, 2007

ok, so I figured out the speed (part a) and as you already know the distance (part b). part a is 100 m/s and part b is 474.105m. Now to find the horizontal and vertical components of its velocity... how would I find this? I see someone said the speed from part a would be the same as the horizontal component..is this correct? How would I find the vertical? This is due tonight, so I need a response asap. Thanks again!

8. Oct 5, 2007

hotvette

In your 2nd post you listed the equation for y = y(t) (assumming I was correct w/ respect to there being a typo). I assume you also have the equation for x = x(t). Here are 2 questions to think about:

1. What can you say about y when the projectile strikes the ground? Can't you find t from this?

2. If you know t when the projectile strikes the ground, how can you use y(t) and x(t) to find the velocity components?

9. Oct 5, 2007