Sutitan
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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
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)
She didnt write anything down, but id assume the kinematic equations
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. Let's 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 that's 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 don't 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 I am not sure.
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. Let's 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 that's 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 don't 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 I am not sure.