Calculating the Final Velocity of a Thrown Ball

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A boy throws a baseball upward and catches it 3.2 seconds later, prompting a discussion on calculating its final velocity. The acceleration due to gravity is acknowledged as negative 9.8 m/s², and the initial velocity is determined using the equation for vertical position over time. The correct approach involves solving the quadratic equation derived from the motion equations, leading to an initial velocity of approximately 31.36 m/s. The discussion emphasizes the importance of correctly applying the equations and keeping track of signs for acceleration and velocity. The final velocity when the ball is caught is negative, indicating it is moving downward.
  • #31
Medgirl314 said:
Great, thanks! The equation I used, or a different one?

The distance versus time equation that I re-summarized in post #25...
 
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  • #32
Okay, I see it now. Thanks! I'll post my answer soon.
 
  • #33
y(t)=y0+viy∗t+(0.5∗ay)∗t2
y(3.2s)=0+0*3.2+(0.5)(-9.8y)*10.24
y(3.2s)=3.2s+(0.5)(9.8y)*10.24sIs that right so far? What's next?
 
  • #34
Medgirl314 said:
y(t)=y0+viy∗t+(0.5∗ay)∗t2
y(3.2s)=0+0*3.2+(0.5)(-9.8y)*10.24
y(3.2s)=3.2s+(0.5)(9.8y)*10.24s


Is that right so far? What's next?

I have no idea what you wrote. Could you please narrate what you are trying to do?
 
  • #35
I'm not really sure. I was trying to plug in everything I know into the equation from post #25. It went horribbly wrong.
 

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