Maximizing Speed: Releasing the Ball from a Cliff

[StephenDoty]

You are standing on a cliff with arm extended. You release a ball.
Which will cause the ball to hit the bottom of the cliff with the greatest speed

Releasing the ball and letting it drop
throwing the ball downward with a velocity of 20m/s
or throwing the ball upward with a velocity of 30m/s

 

[G01]

You must show some thought/work to get help here. What are your thoughts on this problem?

 

[StephenDoty]

If the ball is dropped it accelerates a -9.8m/s
if a ball is thrown downward at 20m/s then the final velocity would be the 20m/s
If a ball was thrown upward at 30m/s then it would have more time to pick up speed

so I think the greatest speed comes from the ball being thrown upward at 30m/s

 

[G01]

You are correct, but I think your reasoning is a little off. If the ball is throw upward with a speed of 30 m/s, when it comes back down to the height of your hand, it'll have a downward velocity of 30m/s. Thus, it's initial velocity on the downward trip is higher. Why does starting at a higher initial velocity mean it'll have a higher final velocity?

Also, the final speed of the ball in #2 is not 20m/s. Isn't it accelerating in this case too?

What you can do for practice is let the height of the cliff be 1, and figure out what the final velocity for each ball would be. It should come out to be #3 like you said, but maybe if you work through each case, it'll help you get a better grasp of what is happening to each ball's velocity as it falls.

 

[kamerling]

If the ball is dropped it accelerates a -9.8m/s
if a ball is thrown downward at 20m/s then the final velocity would be the 20m/s

this does not follow.

If a ball was thrown upward at 30m/s then it would have more time to pick up speed
so I think the greatest speed comes from the ball being thrown upward at 30m/s

It has more time to pick up speed, but you didn't prove it. I wouldn't accept this without a computation.

I assume we can ignore air resistance. This problem is easiest using conservation of energy. In all 3 cases the ball starts with the same potential energy and in all 3 cases the ball ends with the same potential energy. You have to find out in which case the ball has the most kinetic energy at the bottom of the cliff.

 

[StephenDoty]

I was not given the acceleration of any of the cases

If the ball thrown upward has a velocity of 30m/s when it reaches the height of the cliff then it still has gravity acting on it allowing it gain velocity. I could not use 10m for the cliff and solve because I was not given the angle it was thrown. It might have been thrown directly up or not.

I do not know how to even start working with case #2.

thank you for your help. I really appreciate it.
Stephen

 

[StephenDoty]

We have not learned PE=KE even though ke=.5mv^2
and we are not even given the mass

so we can only use the kinimatics equations

 

[PowerIso]

Well, let's think.

We have a start y position called y0 and an finishing y position called y1.

So we know that

v^2 = v^2 o + 2a(y1-y0)

Let's consider the first case:

The starting velocity is 0, so we have

v^2 = 2(-9.8)(-y)

Now consider the second case,

We end up with

v^2 = -20^2 + 2(-9.8)(-y)

Lastly we have

v^2 = 30^2 + 2 -(9.8)(-y)

Out of all these equations, which appears to give you the greatest final velocity?

Edit, I used 1 d motion to keep the idea simple. You can mess with the equations by adding the cosine and sine if you wish to do it in 2d motion. Simply add the cosines and sines where they belong and use arbitrary angles. I'm pretty sure this is the way you should look at this type of problem.