Which ball has the highest velocity just before ground

  • Thread starter Sutitan
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  • #1
My girlfriend recently took a physics exam, and she was confused about why she got one of the problems wrong. She asked me hoping I could clear it up. Heres my attempt

Homework Statement

Three Identical balls are thrown from the top of a building, all with the same initial speed. Ball A is thorwn horizontally, ball B at an angle above the horizontal and ball c at the same angle but below the horizontal. Neglecting air resistance, which ball has the highest speed just before they hit the ground

a) Ball A has the highest
b) Ball B has the highest
c) Ball C has the highest (My girlfriends guess)
d) Balls B and C have same speed
e) Three balls have the same speed (The correct answer)

Homework Equations

She didnt write anything down, but id assume the kinematic equations

The Attempt at a Solution

She wrote "It is C because it is thrown downward, so its not just gravity pulling/pushing it down"

The problem seems pretty straight forward. According to the problem air resistance is neglected, so its essentially like working in a vacuum. So since the ball is constantly under the influence of a gravitational field, it should continue to accelerate (Until it of course is stopped by the ground). Since Ball A is thrown at a horizontal, the y component of the velocity is still zero, so the final velocity is simply [tex]\sqrt{2g(y_{f}-y_{i})}[/tex]. Ball B is thrown above the horizontal (lets assume straight up, since the x component is irrelevant in this scenario). The ball will go up, get to a peak, and then begin to fall. this point will be called [tex]y_{p}[/tex] where [tex]y_{p}[/tex]>[tex]y_{i}[/tex]. since at [tex]y_{p}[/tex], [tex]V_{i}[/tex] = 0, the final velocity is [tex]\sqrt{2g(y_{f}-y_{p})}[/tex]. Ball C is throw down. Lets again ignore the x component. the difference this time is that the is a [tex]V_{i}[/tex]. so for the final velocity, we have [tex]\sqrt{V^{2}_{i}+ 2g(y_{f}-y_{i})}[/tex]

From what I wrote above, it should seem clear that V[tex]_{fballA}[/tex] < V[tex]_{fballb}[/tex], V[tex]_{fballC}[/tex] since both Ball B and Ball C have terms that would increase the final velocity over Ball A's. Ball B had more distance to fall (thus more time to accelerate), and Ball C had an initial velocity. Furthermore, we can prove (im not going to to) the magnitude of the velocity of Ball B initially is the same as the next time it passes y[tex]_{i}[/tex], so essentially, the final velocitys of Ball B and C should be the same.

I think after developing my argument this much, i might have run into a snag. my mind is now thinking that if I throw the ball down, i will lose time spent under the influence of gravity. maybe thats why they all hit the ground with the same velocity. my last argument is kind of a weak one, but here it goes. If I through the ball at the almost the speed of light towards the ground, it would get to the ground at roughly the same speed. I just dont see it being possible for the ball thrown horizontally to be able to go the fast from essentially just dropping it.

Ive managed to confuse myself. I originally believed D, but now im not sure.

Answers and Replies

  • #2
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Think conservation of energy.
  • #3
If you would calculate what the height of the peak was, you would see that
B has the same final velocity as C
The velocity in the x direction is larger for ball A, this makes up for the smaller downwards velocity of A.

The easiest way to see all this is with conservation of energy. In all cases, the loss of potential energy is the same, so the gain of potential energy is also the same. Since the initial speed and initital kinetic energy is the same, the final kinetic energy and thus the speed will also be the same.
  • #4
I see my problem now.

Velocity is a vector. i kept thinking y direction. argh

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