Accel 1-D HW: Solving for Time with Quadratic

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Homework Help Overview

The problem involves a baseball thrown downwards from a height, with the goal of determining the time it takes to reach a friend below. The context is rooted in kinematics and the use of the quadratic formula to solve for time.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the quadratic formula and the implications of obtaining both positive and negative roots. There is confusion about interpreting negative time in the context of the problem.

Discussion Status

Some participants have provided guidance on verifying the results by substituting values back into the original equation. Others are exploring the implications of multiple solutions and how to determine which time is relevant based on the problem's context.

Contextual Notes

Participants note the importance of the problem statement in guiding the interpretation of multiple solutions, particularly in scenarios where a projectile may intersect a given height at different times.

khzak1
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Homework Statement

A baseball is thrown by Joey from the third floor of his home at 20.0m/s [downwards], his friend Timmy is 3.5 m below him.
a. Find how long it takes the ball to reach Timmy.

I used the quadratic formula to solve to time but I am confused about it in my calculations.

Homework Equations



D=V[1]t+1/2(a)(t)^2
1/2(a)(t)^2 + V[1]t + (-D)

The Attempt at a Solution


Once I find the roots using the quadratic equation, how do I find the time? AHH I'm so confused.

20± √468.67
-9.81
 
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khzak1 said:

The Attempt at a Solution


Once I find the roots using the quadratic equation, how do I find the time? AHH I'm so confused.

20± √468.67
-9.81

There are two roots, one is negative, one is positive, does a negative time make any sense as a solution to your problem?
In other words, can Timmy catch the ball before Joey throws it?
 
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So the positive answer relative to time would be
Δt=0.168s approximately

I am not sure if that is the correct answer. Could you confirm it please.
 
khzak1 said:
So the positive answer relative to time would be
Δt=0.168s approximately

I am not sure if that is the correct answer. Could you confirm it please.

It's always to confirm it for yourself, I'm just a stranger on the internet :wink:

Plug t = 0.168 s, v = 20 m/s and a = g into your first equation and see if d comes out as 3.5 m (or close to it due to rounding)
 
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Haha, thanks mate ! :)
 
Something just came to my mind. What if there was a similar question, and I would get a positive integer either way. How would I know which time is correct according to ± ?
 
khzak1 said:
Something just came to my mind. What if there was a similar question, and I would get a positive integer either way. How would I know which time is correct according to ± ?

There should be clues in the problem statement. Imagine a basket ball thrown in a parabola towards the hoop, there will be two positive times where the ball will pass through the same height as the hoop, once on the way up and once on the way down. The second time points are scored (ball passes downward through the hoop) and the first time no points are scored (pall passes upward through the hoop), the question statement will make it clear which time is correct.
 

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