# Homework Help: Help with a Problem Please

1. Sep 18, 2006

### bengaltiger14

This is the problem.

A projectile is being shot from a cannon and must land in a hole 120 meter aways. What angle must the projectile be shot at to land in the hole? Here are some other numbers

VoX=60cos(??) Voy=60sin(???)
VfX=? Vfy=?
Ax=0 Ay=-9.81 m/s^2
t=?? t=??
dx=120m dy=0

2. Sep 18, 2006

### bengaltiger14

Sorry, just read the post that I need to post homework problems elsewhere.

3. Sep 18, 2006

### Delzac

take note that the time for vertical and horizontal motion is the same.

since you know the horrizontal velocity, which is in terms of 60cosZ, can you find time?

4. Sep 18, 2006

### Hootenanny

Staff Emeritus
One is also expected to show ones efforts.

5. Sep 18, 2006

### bengaltiger14

I tried finding t this way dx=Vxt === 120m = 60cosZt

t=120/60cosZ

6. Sep 18, 2006

### Delzac

yeah you are on the right track, try equate the time into your vertical motion equation.

( they land on the ground the same time, the 2 motions)

7. Sep 18, 2006

### bengaltiger14

The answer would be 0 as far as the vertical aspect is concerned right?? I mean the vertical displacement is 0.

8. Sep 18, 2006

### Delzac

correct vertical displacement would be zero

EDITED version

Last edited: Sep 18, 2006
9. Sep 18, 2006

### bengaltiger14

So, would the vertical time be t = sqrt d/VoYt + 1/2a

10. Sep 18, 2006

### Delzac

you can simplify the equation into 0 = t(-Voy + 1/2 g t) ( when solving this type of qns remember to state your sign convention, you don't want the case where you get negative time)

so Voy = 1/2 g t

Last edited: Sep 18, 2006
11. Sep 18, 2006

### bengaltiger14

So t=VoY/2g. But how do I find that if I do not know the angle?? You cant divide 60sin of nothing by 2g can you?

12. Sep 18, 2006

### Delzac

you can equate the 2 equations u have together, the horizontal and vertical, remember the time is the same for both.

13. Sep 18, 2006

### bengaltiger14

Equate them together? So.... t= (120/60cosZ) * (60sinZ/2g)

14. Sep 18, 2006

### Delzac

if you equate in such a way, you have 2 viarables to solve and only one equation to work with.

BTW your equation should be T^2 = ......, since u times the 2 equation together.( but this is not the correct equation to use.)

Last edited: Sep 18, 2006
15. Sep 18, 2006

### bengaltiger14

So that is not the correct way to equate it?? I am really confused here.

16. Sep 18, 2006

### Delzac

T = 120/ (60 cos Z)

T = 2(60 sin Z)/(9.81)

120/(60 cos Z) = 2(60 sin Z)/(9.81)

EDIT : i amended the equations. For the time derieved by vertical motion,

Voy = 1/2 g t
(2 voy)/g = t

t = (2 (60) sin Z) / g

You wrote wrongly the equation for your previous post.

Last edited: Sep 18, 2006
17. Sep 18, 2006

### bengaltiger14

Ok. I'm sorry. I thought I put a equals between the two equations. I did not realize it was a muliplication sign. What do I solve for now?? Do I set the equation equal to 0 and divide -120 by 2(9.81)?

18. Sep 18, 2006

### Delzac

you can cross mutiply. Then isolate your Z, i used double angle formular for this.

BTW this is the first time i use double angle formular for physic. feels strange

19. Sep 18, 2006

### bengaltiger14

So, 120 m * (2*9.81) = 2354s^2. What would 60 sinZ time 60cosZ be???? 3600 sinZ????

20. Sep 18, 2006

### Delzac

this part is wrong, as i have stated about, the equation is

120/(60cos Z) = ((2)(60)sin Z)/ 9.81

thus the correct expression should be 120(9.81) = (2)(60)(60)cos Z sin Z

21. Sep 18, 2006

### Delzac

Then use double angle formular :

$$\sin(2x) = 2 \sin (x) \cos(x) \ = \frac{2 \tan (x)} {1 + \tan^2(x)}$$

22. Sep 18, 2006

### Delzac

Z should be 9.54339 degree

23. Sep 18, 2006

### bengaltiger14

Ok. I am not familiar with double angle forumula. So far I have gotten.

I multiplied and got 1177s^2 = 7200 cosZSinZ

24. Sep 18, 2006

### bengaltiger14

Ok. I see the forumla now. Thanks

25. Sep 18, 2006

### Delzac

now the doucle angle formular states that:

$$\sin(2x) = 2 \sin (x) \cos(x)$$

so we shall express our trigo in this form:

(1177.2)/( 60*60) = 2 sin Z cos Z

2 sin Z cos Z = sin (2Z) as stated by the formular.

There solve from here : sin (2 Z ) = (1177.2)/(60*60)

Last edited: Sep 18, 2006