Calculations involving freefall

AI Thread Summary
The discussion revolves around a physics problem where a baseball pitcher throws a ball vertically upward and catches it after 4.2 seconds. The initial calculations incorrectly used the total time for ascent and descent instead of the time to reach the peak. The correct approach involves using the equation for vertical motion, accounting for the time to the peak, which is half of the total time. The user expresses concern about their understanding of kinematics and seeks advice on effective study strategies for an upcoming test. Emphasis is placed on the importance of time management and consistent practice in mastering the subject.
SPH4UJS
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Homework Statement


A baseball pitcher throws a ball vertically upward and catches it at the same level 4.2s later

a) with what velocity did the pitcher throw the ball?

Homework Equations




The Attempt at a Solution



I put this into givens

t = 4.2 s
displacement = 0 m if the ball goes up and then comes back at the same level I figured it has a displacement of 0

acceleration = -9.81 m/s

Vf = 0 m/s since it goes up then reaches a climax and it was already thrown with an initial velocity that you are trying to find

so I decided to use the vf = vi + a(t)

so.. then I rearranged it so that it was

vf - ( a ) ( t ) = vi

which gave me

-(-9.81)(4.2) = vi

and then I got 41.2 but that's not the right answer :S so I am not sure if its just that I am using a wrong equation or that I have the wrong givens I don't know exactly what I am doing wrong please point me in the right direction.
 
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SPH4UJS said:

Homework Statement


A baseball pitcher throws a ball vertically upward and catches it at the same level 4.2s later

a) with what velocity did the pitcher throw the ball?

Homework Equations




The Attempt at a Solution



I put this into givens

t = 4.2 s
displacement = 0 m if the ball goes up and then comes back at the same level I figured it has a displacement of 0

acceleration = -9.81 m/s

Vf = 0 m/s since it goes up then reaches a climax and it was already thrown with an initial velocity that you are trying to find

so I decided to use the vf = vi + a(t)

so.. then I rearranged it so that it was

vf - ( a ) ( t ) = vi

which gave me

-(-9.81)(4.2) = vi

and then I got 41.2 but that's not the right answer :S so I am not sure if its just that I am using a wrong equation or that I have the wrong givens I don't know exactly what I am doing wrong please point me in the right direction.

With y representing the vertical direction you might want to take a look at

y = y_o + v_o t + \frac{1}{2} a t^2
 
Remember that your using vf as 0 which is at its climax. The time you used is for the total time (up and down) not just the time to get to the climax.
 
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Alright I got the answer thank you very much. Also I have a test tommorow on this whole kinematics unit and so far I know ab out half what would be my best bet? Should I just keep doing questions until I know them all and asking for help if I can't figure it out myself? I really want to do well as I did so poorly on the quiz :S.
 
SPH4UJS said:
Alright I got the answer thank you very much. Also I have a test tommorow on this whole kinematics unit and so far I know ab out half what would be my best bet? Should I just keep doing questions until I know them all and asking for help if I can't figure it out myself? I really want to do well as I did so poorly on the quiz :S.

50 years of experience has taught me that the best bet is NOT to put off studying until the night before the exam!
 
AEM said:
50 years of experience has taught me that the best bet is NOT to put off studying until the night before the exam!

Yeah, it's not the best thing to do :s but I have no idea where my priorities are well..that and my time management skills I have put off doing physics homework until today :S. Hopefully all my questions can be answered on the forums by tomorrow. Its just I feel that I don't have the trained mindset for physics just yet, and actually not learning kinematics scares me for what's to come especially since I want to be an engineer. Btw thank you very much for your help as well as Imperitor.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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