# Homework Help: Help Solve This Linear Equation

1. Oct 9, 2009

### Osbourne_Cox

1. I thought I was doing it right, but Quest said I have to wrong answer. Can some one solve it in steps and produce a final answer so I can compare?

Tan 58=(vi-ayt)/axt

where vi=5.6
ay=-9.8
ax=2.1

solve for t.

I got 0.87 for t.

Thank you.

2. Oct 9, 2009

### rl.bhat

Can you post the actual problem? You have given formula without any information.

3. Oct 9, 2009

### Osbourne_Cox

Initially (at time t = 0) a particle is moving vertically at 5.6 m/s and and horizontally at 0 m/s. The particle accelerates horizontally at 2.1 m/s2 . The acceleration of gravity is 9.8 m/s2 . At what time will the particle be traveling at 58◦ with respect to the horizontal?

4. Oct 9, 2009

### rl.bhat

The horizontal distance traveled = 1/2*ax*t^2
Similarly what is the vertical distance traveled?

5. Oct 9, 2009

d=1/2*ay*t^2

?

6. Oct 9, 2009

### rl.bhat

No.
y = Vo*t - 0.5*g*t^2

7. Oct 9, 2009

### Osbourne_Cox

Okay, so what do I do with those equations? My prof said it should be as easy as Tan 58=(vi-ayt)/axt
and whether you need to break that down more or not, I don't know.

8. Oct 9, 2009

### willem2

What did you do to get t = 0.87?

If you substitute it into the equation you get tan 58 = -1.60, while in reality tan 58 = 1.60,
so I think there's a sign error somewhere

9. Oct 9, 2009

### Osbourne_Cox

vi-at=tan 58 (at)
-(-9.8)t=tan58(2.1)t-5.6
9.8t=3.36t-5.6
6.44t=-5.6
t=0.87

but time cant be negative..?

10. Oct 9, 2009

### rl.bhat

This formula is wrong. It should be
Tanθ = (2vi -ayt)/axt

11. Oct 9, 2009

### Osbourne_Cox

so i get -1.74=t

do we just take the absolute value for the time? or have I made another mistake

12. Oct 9, 2009

### willem2

you replace a_x and a_y both by a here wich is confusing.
You made a sign error when moving 5.6 to the right side.
Another problem that i can see now is that when replacing $v_y$ with$(v_i-a_yt)$ to get your original equation this minus sign already accounted for the fact that gravity would be in the negative y direction. So you should substitute g = 9.8 here, and not g = -9.8
If your prof really gave you that exact equation with $a_y$ instead of g that would
be highly misleading, if not wrong.

13. Oct 9, 2009

### Osbourne_Cox

Yes, and as a result I have spent a week being confused by this problem...

14. Oct 9, 2009

### rl.bhat

Instead of giving the answer, it will be better to show the actual calculations by substituting the values. You can rewrite the expression as
tan58 = Vo/ax*t - ay/ax. Now solve for t.

15. Oct 9, 2009

### willem2

Where do you get the 2 from? The equation comes from $$tan(58) = \frac { v_y} {v_x}$$

you just substitute $v_y = v_i - gt$ and $v_x = a_x t$ into it

16. Oct 9, 2009

### Osbourne_Cox

okay, so using that I get t=0.426s

17. Oct 9, 2009

### rl.bhat

In the problem, the time at which position of the particle makes an angle 58 degrees to the horizontal, is required.

18. Oct 10, 2009

### willem2

That is not how I interpret this sentence

19. Oct 10, 2009

### rl.bhat

In that case it should be " the net velocity in the direction 58 degrees to the horizontal." And I feel, in the problem position is more appropriate than the velocity.

20. Oct 10, 2009

### Osbourne_Cox

Instead of giving the answer, it will be better to show the actual calculations by substituting the values. You can rewrite the expression as
tan58 = Vo/ax*t - ay/ax. Now solve for t.

This formula gave me the right answer.