# Projectile motion cannon problem

1. Aug 29, 2007

### EvanQ

1. The problem statement, all variables and given/known data

A cannon, located 60.0m from the base of a vertical 25.0m tall cliff, shoots a 15 kg shell at 43.0degrees above the horizontal toward the cliff.

What must the minimum muzzle velocity be for the shell to clear the top of the cliff?

The ground at the top of the cliff is level, with a constant elevation of 25.0 m above the cannon. Under the conditions of part A, how far does the shell land past the edge of the cliff?

2. Relevant equations

2-dimensional motion equations such as s=ut+(1/2)at^2 etc. also tried using the range formula 2U/g sin(theta) cos(theta).

3. The attempt at a solution

i've been stuck on this problem for hours and feel like i've exhausted all of the options that i understand / am aware of. i initially tried to get a reading of time by calculating how long it would take for the ball to drop the 25m, and came out with approx 2.259, then used s=ut+(1/2)at^2 to calculate that the vertical velocity needed to make the 25m was 22.136m/s. But then after rearranging that to find the actual muzzle velocity, i think i found that that was insufficient to make the 60metres. so then i tried tackling it using that range formula i listed, and came out with a muzzle velocity of 25.23m/s, which out of frustration i entered to find i was incorrect. i have 2 more submission attempts and i am really confused as to how to attack this problem. any help would be greatly appreciated.

-evan.

2. Aug 29, 2007

### learningphysics

Can you write down the equations for horizontal and vertical displacement in terms of t?

3. Aug 29, 2007

### EvanQ

i don't know how to >.<

4. Aug 29, 2007

### learningphysics

Deal with the horizontal aspect and vertical aspect separately...

So deal with the horizontal aspect firstly.

There is no horizontal acceleration.... so the horizontal component of the velocity is constant... what is this horizontal velocity? Write it in terms of v and $$\theta$$. You need a little trigonometry here. So then what is the horizontal displacement over time... remember that velocity is constant in the horizontal direction...

5. Aug 29, 2007

### learningphysics

Right when the cannon is fired, what's the horizontal velocity? and what's the vertical velocity?

Write these in terms of v and $$\theta$$

6. Aug 29, 2007

### EvanQ

horizontal velocity would be Vcos43?
so the horizontal displacement over time would be t=60/vcos43

and the vertical velocity would be Vsin43.

so would i then factor in that vertically the canonball has to go a displacement of 25m, with an acceleration of -9.8m/s, and sub 60/vcos43 in for t?

7. Aug 29, 2007

### learningphysics

Yup. Sounds good to me. Solve for v, and that gives your answer.

Last edited: Aug 29, 2007
8. Aug 29, 2007

### EvanQ

ok thanks, 1 last problem though.

after subbing into s=vt+.5at^2 i got:
25=(vsin43)(60/vcos43)+(-4.9)(60/vcos43)

and got all the way through rearranging and only doing simple multiplications till i got to:
25=60sin43/cos43 - 294/vcos43

then after using my calculator to do the rest, came up with a v value of 2.5586 which is obviously wrong. my calculator is in RAD but seems to be acting weird... should i change it to DEG or GRAD, or have i gone wrong in my rearranging?

9. Aug 30, 2007

### learningphysics

This should be:
25=(vsin43)(60/vcos43)+(-4.9)[(60/vcos43)^2]

You forgot to square...

10. Aug 30, 2007

### learningphysics

You're given 43 degrees... so you should be using DEG for your calculator.

11. Aug 30, 2007

### EvanQ

alright thanks, i re-arranged again and think i have come up with the right answer.

this is for an online assignment and i get 3 attempts at submitting, i've already used 2 so this is my last chance.

i've got 43.84m/s at the moment.
is this right? it would be greatly appreciated if you could let me know, if it's wrong just say and i will keep on trying.

12. Aug 30, 2007

### learningphysics

I'm not getting that... keep trying. I'll check my work also.

13. Aug 30, 2007

### EvanQ

i'm doing this with the working:
25=(vsin43)(60/vcos43)+(-4.9)(60/vcos43)
25 - (-4.9)(60/vcos43) = (vsin43)(60/vcos43)
25 - (-4.9) = (vsin43)
29.9/sin43 = v
v = 43.84m/s

where have i gone wrong?
i keep trying different methods of re-arranging and coming up with different answers every time... not too sure when i can cancel things out etc when dealing with trig sorry.

14. Aug 30, 2007

### learningphysics

You're missing the square again... see my post earlier... post #9

15. Aug 30, 2007

### EvanQ

ahhh thanks, sorry missed that post.
have to go to work now, but if i just put the square in i should be ok?
it's not due until around this time tomorrow so i'll still put my answer here before putting it in just to check...

thanks a lot for your help.

16. Aug 30, 2007

### learningphysics

That's cool. I'll check. Be careful with your algebra... the square was the main mistake but your algebra also had errors...

17. Aug 30, 2007

### EvanQ

i'll go through my algebra with this one in case i've stuffed it again.

25=(vsin43)(60/vcos43)+(-4.9)(60/vcos43)^2
25=(60vsin43/vcos43)+(-4.9)(3600/(v^2)(cos43)^2)
25=(60sin43/cos43)-(17640/(v^2)(cos43)^2)
30.95=(17640/(v^2)(cos43)^2)
569.935=(v^2)(cos43)^2)
(v^2)=1065.54
v=32.6426m/s

pleeeeeeeeeeeeeeeeeeease be right.

18. Aug 30, 2007

### learningphysics

Yup. That's exactly what I get. Let me know if the answer goes through.

19. Aug 30, 2007

### EvanQ

yep correct :)
thanks too much for that.

now onto part two of the question lolol.
The ground at the top of the cliff is level, with a constant elevation of 25.0 m above the cannon. Under the conditions of part A, how far does the shell land past the edge of the cliff?

i'll let you know how i go.

20. Aug 30, 2007

Cool. :)