Concept of 3 balls falling from building at different angles

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When three balls are thrown from a building at different angles—upwards, horizontally, and downwards—they will all reach the ground with the same final velocity if launched with the same initial speed and neglecting friction. Calculations show that the vertical launch reaches a height of 55.102 meters before descending, while both the horizontal and downward launches yield a final speed of 32.863 m/s. The discussion highlights that the upward and downward cases are essentially the same due to energy conservation principles. The horizontal launch's complexity arises from its combination of vertical and horizontal components, yet it ultimately aligns with the same final speed. This problem illustrates fundamental physics concepts, including gravitational potential energy and kinetic energy.
mrbrownstone
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If a ball is thrown upwards, another straight out, and another downwards, which will reach the ground with the highest velocity? After thinking about this for a while my conclusion is that they all will have the same velocity since they all have the same initial gravitational potential energy. Am I on the right track?
 
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Yes. Assuming they all are given the same initial speed, you are correct.
 
Thank you very much for the input. I'm glad I found this forum, it seems to be an excellent resource.
 
No ha ha
It depends on what velocitys they are being launched at, and I am assuming that you are neglecting friction.
So let's say they all have the same initial velocities and start at 50 meters up. The starting velocities are 10m/s.
Vertical Launch: So we must first find out how high it goes before it stops and then starts heading back down to earth. I have calculated using the formula H(height)= Vi(initial veritcal velocity)T(time)+1/2(-9.8[acceleration due to gravity])(T2) here it is without the writing H=Vi*T-1/2*(9.8)*T2... I got that it reached 55.102 meters. The using PE=KE i got that it will have a speed of 32.863 m/s

Horizontal launch: So here it gets complicated but i will give you two answers, first its overall velocity cause it will have both an X-axis component, its initial veloctiy, and a Y-axis component. The second is just the Y-axis part. So I use the equation 2HG(acceleration due to gravity)= Vfinal2-Vinitial2 to find the final vertical velocity. And i get 31.305 m/s. Thats just the Y-axis componet. the overall speed is 32.863 m/s

Straight down launch: I will use the same method as Horizontal launch to find the final velocity here. (use same formula) and i get that the final speed is 32.863...

Im a little bit shocked at what I am getting but all my numbers are correct. So it seems that yes they will all hit the ground with the same speed... That is that they have all been launched at the same initial speed

This is kinda cool and weird to me ha ha, Hope this answers your question
 
:smile: Nice
 
One thing that could have been noted, that for the ball thrown upwards, eventually it returns back to the point from where it was thrown from, with the same speed, but now downwards, so it should be clear that the upwards and downwards case is the same.

The horizontal case isn't so obvious unless you take into account that the total energy, KE + GPE is the same as the upwards and downwards case. This assumes that the object's speed isn't fast enough that curvature of the Earth's surface would be significant.
 
Jeff Reid said:
This assumes that the object's speed isn't fast enough that curvature of the Earth's surface would be significant.
Uh, really?
 
cesiumfrog said:
Uh, really?

well yeah, if you threw it hard enough it would never hit the ground as it would reach escape velocity (I know its not being thrown in the most efficient direction to exit the Earth's pull but it still has a Z component meaning it can be done with enough force)

I think you are looking at a lot of decimal ponts before it's a factor in the real world, but it's worth mentioning
 
Wow... at first I was very confused, now I get it. Interesting problem!
 

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