Kinetic and gravitational potential energy and velocity

[pbonnie]

Homework Statement

A watermelon with a mass of 2.0kg falls out of a tree house that is 5.4m above the ground. What is the speed of the watermelon just before it hits the ground?

Homework Equations

Δh = (vi^2 - vf^2)/2g
Eg = mgΔh
Ek = 1/2mv^2

The Attempt at a Solution

I tried two different attempts at this one. The first involves rearranging the first equation to solve for the final velocity, since the watermelon is falling out of the tree fort I assume the initial velocity is 0.

On my actual answer I would show the calculations, but I got the answer v = 10.3 m/s

My other attempt (which I'm not sure actually applies to this) is to figure out the gravitational potential energy, and then assume that because the watermelon is hitting the ground that all of that energy would be converted into kinetic energy, so then I could solve for velocity.
Doing this, I got the same answer. So I could use either method?

 

[Doc Al]

Doing this, I got the same answer. So I could use either method?

Yes, both methods are equivalent. If you express them mathematically, you'll end up with the same final equation using either method. (Try it and see!)

 

[Dick]

Homework Statement

A watermelon with a mass of 2.0kg falls out of a tree house that is 5.4m above the ground. What is the speed of the watermelon just before it hits the ground?

Homework Equations

Δh = (vi^2 - vf^2)/2g
Eg = mgΔh
Ek = 1/2mv^2

The Attempt at a Solution

I tried two different attempts at this one. The first involves rearranging the first equation to solve for the final velocity, since the watermelon is falling out of the tree fort I assume the initial velocity is 0.

On my actual answer I would show the calculations, but I got the answer v = 10.3 m/s

My other attempt (which I'm not sure actually applies to this) is to figure out the gravitational potential energy, and then assume that because the watermelon is hitting the ground that all of that energy would be converted into kinetic energy, so then I could solve for velocity.
Doing this, I got the same answer. So I could use either method?

Sure you can. The first equation can be derived from the energy relation. Use whatever seems easiest.

 

[pbonnie]

Great thank you :)

Also, I just wanted to double check my solution to the second part of that question.
B) A cantaloupe with a mass of 0.45kg falls out the other side of the tree house. It hits a branch at a speed of 6.3m/s. How high is the tree branch from the ground?

I said:
m = 0.45 kg vi = 0 vf = 6.3 m/s ∆htree = 5.4m ∆hbranch = ?
Find the height of the branch from where it dropped to where it hit the branch, then subtract that from the total height.
∆h= (vi^2-vf^2)/2g ∆h= (0-〖(6.3m/s)〗^2)/(2(9.81m/s^2)) ∆h = 2.0 m (from where it dropped)
∆h = 5.4m – 2.0m = 3.4m
The tree branch is 3.4 m above the ground.

Is this correct?

 

[Dick]

Sounds just fine to me.

 

[Doc Al]

The tree branch is 3.4 m above the ground.

Is this correct?

Yes. Looks good.

 

[pbonnie]

Great, thank you both very much!