TBR Physics question

[Mwebs]

From the 25 qs after chapter 1 (Kinematics)

3 canon balls are launched from the edge of a cliff 100 meters above a canyon floor. They are all launched with the same initial speed of 50 m/s but at different angles.
A cannon is located on the edge of a cliff 100 m above a flat, sandy canyon floor. Three cannon balls are launched from the cannon with the same initial speed of 50 m/s, but at different angles. The cannon balls are identical except for their color. A red ball is launched horizontally, a white ball is launched at a 30° angle above the horizontal, and a blue ball
is launched at a 60° angle above the horizontal. In all questions you may ignore air resistance and the length of the cannon, unless otherwise stated.

Which of the three cannon balls hits the ground with the GREATEST kinetic energy?
A. Red
B. White
C. Blue
D. They all hit the ground with the same kinetic
energy.

I understand why the answer is D based on conservation of energy.
KEi + PEi = KEf + PEf
Since KEi is 0 and PEi is the same for all, and PEf is the same for all, then KEf must be equal.
What I don't understand is how this can be since this would mean they all have the same velocity???
Wouldn't the ball that went higher (launched at angle of 60) be moving faster already when it gets back to the initial height and thus be moving faster when it hits the ground? I'm really confused...

---

Members don't see this ad.

 

[clothcut]

If you look at the formula you have for energy conservation, PE = mgh. Regardless of their paths, what do each of the balls have in common INITIALLY? mass AND height. They are being shot from the same height and mass, with g as a constant. so all their K.E. will be the same.

To be more detailed, consider this scenario... the ball that is shot at 60 degrees reaches its apex and falls back down, right? But by the time it reaches its original 'h,' the differences in velocities cancel out (because upward v and downward v are equal and opposite) and it is as if it started from that point all along... hope that makes sense.

 

[HotHamH2O]

From the 25 qs after chapter 1 (Kinematics)

3 canon balls are launched from the edge of a cliff 100 meters above a canyon floor. They are all launched with the same initial speed of 50 m/s but at different angles.
A cannon is located on the edge of a cliff 100 m above a flat, sandy canyon floor. Three cannon balls are launched from the cannon with the same initial speed of 50 m/s, but at different angles. The cannon balls are identical except for their color. A red ball is launched horizontally, a white ball is launched at a 30° angle above the horizontal, and a blue ball
is launched at a 60° angle above the horizontal. In all questions you may ignore air resistance and the length of the cannon, unless otherwise stated.

Which of the three cannon balls hits the ground with the GREATEST kinetic energy?
A. Red
B. White
C. Blue
D. They all hit the ground with the same kinetic
energy.

I understand why the answer is D based on conservation of energy.
KEi + PEi = KEf + PEf
Since KEi is 0 and PEi is the same for all, and PEf is the same for all, then KEf must be equal.
What I don't understand is how this can be since this would mean they all have the same velocity???
Wouldn't the ball that went higher (launched at angle of 60) be moving faster already when it gets back to the initial height and thus be moving faster when it hits the ground? I'm really confused...

No. It would be traveling at the exact same speed it was when it left that initial height.

 

[Mwebs]

Well I was thinking about it last night and I get it now (really tricky question if you really think about it, not if you just look at cons of energy, then its simple)
Basically, the y component of V WILL be higher for the 60 degree ball! But KE takes velocity into account, meaning its not just talking about V in the y direction, so any extra PE gained by the work done on the 60 degree ball to raise it higher in the arc, is also done on the 30, and the horizontal, just to move them in the x direction. So all 3 balls have the same velocities when they hit the ground (albeit in different directions).

Waaay too much thinking for a conservation of energy problem...

 

[HotHamH2O]

Well I was thinking about it last night and I get it now (really tricky question if you really think about it, not if you just look at cons of energy, then its simple)
Basically, the y component of V WILL be higher for the 60 degree ball! But KE takes velocity into account, meaning its not just talking about V in the y direction, so any extra PE gained by the work done on the 60 degree ball to raise it higher in the arc, is also done on the 30, and the horizontal, just to move them in the x direction. So all 3 balls have the same velocities when they hit the ground (albeit in different directions).

Waaay too much thinking for a conservation of energy problem...

I think the point of this problem is they want to show you that angle of initial velocity is irrelevant to the final velocity in projectile motion.

½m • v_i² = ½m • v_f²

---> v_i² = v_f²

---> so looking at KE_f the velocities are equal for all of the balls because they all have the same initial velocity and initial velocity is equal to final velocity in the absence of wind resistance and same initial height making all KE_f equal.

I hope I didn't screw any of this up, but I'm pretty confident that this is the way to look at it.

 

[radish2]

.

 

[Mwebs]

Yeah I realized that, thanks. That's what I didn't realize at first.

Thanks