Mechanical Energy of two Cannonballs fired at two different angles

[Mustard]

Homework Statement

A 20.0-kg cannonball is fired from a cannon with muzzle speed of 1,000 m/s at an angle of 37.0 degrees with the horizontal. A second ball is fired at an angle of 90.0 degrees. Use the isolated system model to find (a) the maximum height reached by each ball and (b) the total mechanical energy of the ball–Earth system at the maximum height for each ball. Let y=0 at the cannon.

Relevant Equations

ME = PE + KE

I solved for both heights: h1 = (1.85x10^4)m, h2 = (5.10 x 10^4)m

Question is knowing ME = PE + KE for ball at h1 (angled 37 degrees), is it safe to assume there would be no KE as velocity would equal zero at the highest point? So the mechanical energy = mgh ? Would it be the same for ball at h2?

 

[etotheipi]

What are the horizontal components of velocity of each ball at the highest points in their trajectory? Hint: Draw a diagram of each trajectory and draw on the velocity vector at the highest point.

 

[Mustard]

What are the horizontal components of velocity of each ball at the highest points in their trajectory? Hint: Draw a diagram of each trajectory and draw on the velocity vector at the highest point.

By horizontal components do you mean Vx? For the first it would be Vx=1000(cos37degrees), 2nd ball would not have a Vx component?

 

[etotheipi]

By horizontal components do you mean Vx? For the first it would be Vx=1000(cos37degrees), 2nd ball would not have a Vx component?

Right, yes. So what's the kinetic energy of each at the highest point in the trajectory?

 

[Mustard]

Right, yes. So what's the kinetic energy of each at the highest point in the trajectory?

Hmmm, conceptually...I am having a hard time grasping why wouldn't kinetic energy be equal zero at the highest point since velocity would be zero?

But for the first ball, it won't have kinetic energy in the y-direction but it would have kinetic energy in the x-direction?

KEx=0.5(20)(Vx)^2​

For the 2nd ball, if we continue this train of thought it would be no KE in x-direction and KE in y-direction:

KEy=0.5(20)(1000)^2​

 

[etotheipi]

The kinetic energy of a body is defined as $T = \frac{1}{2}m(\vec{v} \cdot \vec{v}) = \frac{1}{2}mv^2 = \frac{1}{2}m({v_x}^2 + {v_y}^{2}) = \frac{1}{2}m{v_x}^2 + \frac{1}{2}m{v_y}^2$.

You can then in a sense treat the kinetic energy in both directions separately. All that you need to do is find $v_x$ and $v_y$ at the top of the motion. You already know that $v_y = 0$ at the top of the trajectory since you used this fact to determine the heights. And in post #3 you worked out the horizontal components, $v_x$.

So what are the kinetic energies of each?

 

[Mustard]

The kinetic energy of a body is defined as $T = \frac{1}{2}m(\vec{v} \cdot \vec{v}) = \frac{1}{2}mv^2 = \frac{1}{2}m({v_x}^2 + {v_y}^{2}) = \frac{1}{2}m{v_x}^2 + \frac{1}{2}m{v_y}^2$.

You can then in a sense treat the kinetic energy in both directions separately. All that you need to do is find $v_x$ and $v_y$ at the top of the motion. You already know that $v_y = 0$ at the top of the trajectory since you used this fact to determine the heights. And in post #3 you worked out the horizontal components, $v_x$.

So what are the kinetic energies of each?

Ohhh! So KE for ball 1 = 0.5(20)(Vx)^2 , 2nd ball would not have KE. So told solve for ME for ball one would be:

ME = KE
*No PE since there is KE?*​

Solving for ME for ball two would be:

ME = PE​

 

[etotheipi]

So told solve for ME for ball one would be:

ME = KE
*No PE since there is KE?*​

The Earth-ball system definitely still has PE! It's just it also has KE too.

 

[Mustard]

The Earth-ball system definitely still has PE! It's just it also has KE too.

Hmmm, I think I got it! ME for both are equal around 10,000,000 J. Thank you very much, I appreciate you!

 

[PeroK]

Hmmm, I think I got it! ME for both are equal around 10,000,000 J. Thank you very much, I appreciate you!

Note that mechanical energy is conserved in these scenarios, which means it is constant over time, and depends only on the mass and initial muzzle velocity.