How High Does a Cannonball Go When Fired Straight Up at 28m/s?

[mayhemlikeme]

Homework Statement

a cannon is fired out of a cannon at a velocity of 28m/s and at an angle of 90°. Straight up. Assume it starts at a height of 0, figure out the max height it will reach. Air resistance is ignored.

Homework Equations

The Attempt at a Solution

I'm not 100% on this but I think with conservation you can use the equation: mgh= 1/2mv^2
I just need help setting it up I'm usually pretty good working it out. Thanks in advance.

 

[Dick]

Homework Statement

a cannon is fired out of a cannon at a velocity of 28m/s and at an angle of 90°. Straight up. Assume it starts at a height of 0, figure out the max height it will reach. Air resistance is ignored.

Homework Equations

The Attempt at a Solution

I'm not 100% on this but I think with conservation you can use the equation: mgh= 1/2mv^2
I just need help setting it up I'm usually pretty good working it out. Thanks in advance.

You've already set it up correctly. Cancel the m and solve for h.

 

[mayhemlikeme]

You've already set it up correctly. Cancel the m and solve for h.

How do you set it up without knowing mass?

 

[Dick]

How do you set it up without knowing mass?

Put the mass equal to 1kg. Solve it. Put the mass equal to 2kg. Solve it. You'll get the same answer. The mass doesn't matter. It cancels.

 

[Nickie]

Hello! It seems like you are trying to calculate the maximum height reached by a cannonball when fired straight up with a velocity of 28m/s and ignoring air resistance. To solve this problem, we can use the principle of conservation of energy, which states that energy cannot be created or destroyed, but can only be transferred from one form to another. In this case, we can consider the initial kinetic energy of the cannonball (1/2mv^2) to be converted into potential energy at the highest point of its trajectory (mgh).

To set up the equation, we can start by identifying the variables given in the problem:

- Velocity (v) = 28 m/s
- Angle (θ) = 90°
- Height (h) = 0
- Mass (m) = unknown

Next, we can use the equation you mentioned, mgh = 1/2mv^2, and plug in the given values:

- mgh = 1/2mv^2
- (m)(9.8)(0) = 1/2(m)(28)^2
- 0 = 1/2(m)(784)
- 0 = 392m

From this, we can see that the mass (m) cancels out and we are left with 0 = 0. This means that the mass of the cannonball does not affect the maximum height reached and therefore, the maximum height is solely determined by the initial velocity and angle.

To solve for the maximum height, we can rearrange the equation to solve for h:

- mgh = 1/2mv^2
- h = 1/2v^2/g
- h = 1/2(28)^2/9.8
- h = 392/9.8
- h = 40 meters

Therefore, the maximum height reached by the cannonball is 40 meters. I hope this helps! Let me know if you have any further questions.