Need help on another freefall problem.

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Homework Help Overview

The discussion revolves around a freefall problem involving a foul ball hit straight up with an initial speed of 35 m/s. The original poster has calculated the maximum height reached by the ball but seeks assistance in determining the total time the ball is in the air after being caught at a height of 4 meters above the bat.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations to find the time of flight but questions the accuracy of their calculations and the values used. Participants raise concerns about variable naming conventions and the initial conditions of the problem, suggesting that the original poster may have misinterpreted the variables involved.

Discussion Status

Participants are actively engaging in clarifying the problem setup and addressing potential misunderstandings in the original poster's approach. There is a focus on ensuring correct variable usage and understanding the relationship between the maximum height and the subsequent motion of the ball.

Contextual Notes

The original poster has been provided with an answer by their teacher, which adds a layer of complexity to the discussion as they seek to reconcile their calculations with this provided solution. There is an emphasis on the need for clarity regarding the definitions of displacement and initial conditions in the context of the problem.

Bensky
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Need help on another freefall problem. :(

Homework Statement


A foul ball is hit approximately straight up into the air with a speed of 35 m/s.
a) How high does it go (above the bat): Already figured this out, 62.5m. The initial velocity in this problem was 35 m/s.
b) If it clears the backstop on the way up but gets caught and stuck in the backstop 4.00 meters above the bat on the way back down, how long is it in the air? (need help on this one)

variables:
t=?
g=-9.8 m/s^2
x=4 m
x0=62.5m
v0=0 m/s
v=0 m/s

Homework Equations


x=x0+v0t+(1/2)at^2


The Attempt at a Solution


x=x0+v0t+(1/2)at^2
4=62.5+(1/2)(-9.8)(t^2)
-58.5=-4.9t^2
t^2=11.93877551
t=3.455253321

Now, the correct answer to this problem is supposed to be 7.03s according to my teacher (he gave us the answers to check our work), but I obviously did something wrong and can't figure out what it is. I'm thinking maybe x isn't 4m, but I'm not sure. Any help is appreciated, thanks. :)
 
Last edited:
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First thing is first, there is no x in this problem. It may sound stupid..but do NOT use x as avariable when dealing with vertical displacement. USE y.

Secondly..4m IS displacement...why do you have 62.5 plugged in for v_0 ?

Casey
 
Saladsamurai said:
First thing is first, there is no x in this problem. It may sound stupid..but do NOT use x as avariable when dealing with vertical displacement. USE y.

Secondly..4m IS displacement...why do you have 62.5 plugged in for v_0 ?

Casey

62.5 isn't plugged into v0, I have 0 plugged into that. :confused: Did you mean y0?
 
Oh. Sorry. But now that makes it doubly incorrect. Is v_0=0? Hint:no.

What does the fact that the MAX HEIGHT that the ball travels is 62.5m have to do with this part of the problem? Hint: nothing.

Casey

Draw a Diagram and see what you can deduce from it and get back to me.

What is v_0...y_0...y_f?...
 

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