Projectile motion cannon problem

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Homework Help Overview

The problem involves a cannon firing a shell at an angle towards a cliff, requiring the determination of the minimum muzzle velocity needed for the shell to clear the cliff's height of 25.0 m, located 60.0 m away from the cannon. The discussion also touches on how far the shell lands past the cliff after clearing it.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of kinematic equations for both horizontal and vertical motion, questioning how to separate the components of motion and apply trigonometric relationships. There are attempts to calculate time of flight and vertical displacement, with some participants expressing confusion over their calculations and the use of their calculators.

Discussion Status

Several participants have provided guidance on breaking down the problem into horizontal and vertical components. There is ongoing exploration of algebraic rearrangements and the impact of squaring terms in the equations. Some participants have reported their calculated values and are seeking confirmation of their results, indicating a collaborative effort to clarify the problem-solving process.

Contextual Notes

Participants mention constraints related to submission attempts for an online assignment, which adds urgency to their discussions. There is also a focus on ensuring the correct use of degrees in calculations, as the problem involves angles.

EvanQ
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Homework Statement



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?


Homework 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).

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.
 
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Can you write down the equations for horizontal and vertical displacement in terms of t?
 
i don't know how to >.<
 
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...
 
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
 
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?
 
EvanQ said:
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?

Yup. Sounds good to me. Solve for v, and that gives your answer.
 
Last edited:
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?
 
EvanQ said:
ok thanks, 1 last problem though.

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

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

You forgot to square...
 
  • #10
EvanQ said:
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?

You're given 43 degrees... so you should be using DEG for your calculator.
 
  • #11
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
EvanQ said:
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.

I'm not getting that... keep trying. I'll check my work also.
 
  • #13
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
EvanQ said:
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.

You're missing the square again... see my post earlier... post #9
 
  • #15
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
EvanQ said:
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.

That's cool. I'll check. Be careful with your algebra... the square was the main mistake but your algebra also had errors...
 
  • #17
alright had another go.
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
Yup. That's exactly what I get. Let me know if the answer goes through.
 
  • #19
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
EvanQ said:
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.

Cool. :)
 
  • #21
was 0m, got it correct. thanks again.
 
  • #22
EvanQ said:
was 0m, got it correct. thanks again.

No prob.
 

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