Projectile motion of a fired cannonball

  • Thread starter roam
  • Start date
  • #1
1,266
11

Homework Statement



A cannonball is fired with an initial speed of 90.0 m/s at an angle 45.0 ° above the horizontal (see diagram). The cannonball strikes point A on top of a cliff 8.7 s seconds after being fired. Ignore air resistance in this problem.

How high is the cliff? (correct answer is: 183 meters)


The Attempt at a Solution



[tex]v_{iy}=90 sin(45)=63.63[/tex]

[tex]y=v_{iy}t+ \frac{1}{2}(a)t^2=63.63 \times 8.7 + 0.5 (9.81)\times 8.7^2 = 924.8[/tex]

But why is this wrong? :confused:

P.S. I also know that the the maximum height the cannonball reaches is 206.42 and I have already calculated that point A is 553.66 meters horizontally from the cannon. Not sure if this helps...
 

Answers and Replies

  • #2
66
0
you used the formula s = ut + 0.5at^2, the acceleration in this case is in the opposite direction of the motion of the cannonball, so effectively it is decellerating it. So instead of 9.81ms^-2 you need to plug in -9.81ms^-2.
 
  • #3
1,266
11
I also have another relevant question:

Calculate the speed of the cannonball just before it hits A.

I used the formula [tex]v=v_i + at = 90 + (\pm 9.81)8.7[/tex]

I have tried both 9.81ms-2 and -9.81ms-2, but neither seem to give the correct answer. Am I using the wrong formula or something? :rolleyes:
 
  • #4
66
0
I also have another relevant question:

Calculate the speed of the cannonball just before it hits A.

I used the formula [tex]v=v_i + at = 90 + (\pm 9.81)8.7[/tex]

With that formula you are only finding the vertical component of the speed of the cannonball. Remember that the ball is moving in the horizontal plane aswell.

Also, make sure that you use the correct sign infront of the 9.81ms-2. If gravity is acting in favor of the direction of the object, it's accellerating it, otherwise it's decellerating it and hence must be accounted for differently.
 
Last edited:
  • #5
1,266
11
With that formula you are only finding the vertical component of the speed of the cannonball. Remember that the ball is moving in the horizontal plane aswell.

Also, make sure that you use the correct sign infront of the 9.81 9.81ms-2. If gravity is acting in favor of the direction of the object, it's accellerating it, otherwise it's decellerating it and hence must be accounted for differently.

So, what formula would you propose? :confused:
 
  • #6
66
0
Once you find the horizontal velocity, you need to add the two together (vectorially) to get the resultant velocity which is probably what you're being asked for.
 
  • #7
1,266
11
Once you find the horizontal velocity, you need to add the two together (vectorially) to get the resultant velocity which is probably what you're being asked for.

I didn't know that the formula v=vi + at is only for the vertical velocity. Is there a formula that gives us the resultant velocity?

Does the formula v=xt give the horizontal velocity? If so then 553.66x8.7=1816.8, and if we add the two together we get 1816.8+4.7=1821.45. But this is very different from the correct answer (183 meters)!
 

Related Threads on Projectile motion of a fired cannonball

Replies
10
Views
5K
Replies
3
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
7
Views
11K
  • Last Post
Replies
3
Views
1K
Replies
1
Views
9K
Replies
5
Views
418
Replies
5
Views
7K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
5K
Top