Ball thrown up, find max height - Given one velocity in midflight

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Homework Help Overview

The problem involves a ball thrown upwards from a point near the sidewalk, with a specific scenario where an occupant of a building observes the ball passing her window at a certain speed. The goal is to determine the maximum height reached by the ball, given the height of the window and the speed at which the ball is observed.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to find the initial velocity and maximum height. There are attempts to clarify the calculations and assumptions made regarding the ball's motion and the observer's position.

Discussion Status

Some participants have provided guidance on re-evaluating the calculations and considering the problem from different perspectives, such as starting from the observer's height. There is an ongoing exploration of the correct approach to solve for the maximum height, with no explicit consensus reached yet.

Contextual Notes

Participants note the importance of correctly applying kinematic equations and the potential confusion arising from the height of the observer and the speed of the ball at that point. There is mention of neglecting air resistance and using a specific value for gravitational acceleration.

ichivictus
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Homework Statement



A ball is thrown upwards from a point near the sidewalk next to a tall office building. At some later time an occupant of the building, exactly 5.1 meters above the point from which the ball was released, sees it pass her window with a speed of 2.9 m/s.

Calculate the maximum height reached by the ball.

You may neglect air resistance. Take the acceleration due to gravity to be 9.82 m/s^2


Homework Equations


Kinematic Equations

y = Vot + (1/2)at2
Vf = Vo + at
Vf2 = Vo2 + 2ay
y = t * [ (Vo + Vf) / 2 ]


The Attempt at a Solution



Using Kinematic equation Vf^2=Vo^2+2ay I solve for Vo
(2.9)^2=Vo^2 - 19.64*5.1
Vo = 10.01 m/s

To find max height I set Vf to 0 and solve for y using the same acceleration and the newly found Vo.

0 = 10.01^2 - 19.64y
y=5.1 m

Obviously this doesn't work since 5.1m is where the person at the window sees the ball traveling with velocity. Any help would be great, I tried using the other kinematic equations but it got too messy too quickly.
 
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Your value for vo does not look right. I think you forgot to add the (2.9)^2 term.
 
nasu said:
Your value for vo does not look right. I think you forgot to add the (2.9)^2 term.

That must have been it. I just redid the problem and received an answer a little greater than 12 meters. Not sure where I went wrong exactly, but no worries.

Thanks :)
 
Here's how I would look at it. Start from the window.

The ball is traveling upwards at 2.9 m/s, so how high does it go from there?

Just add that answer to the starting point of 5.1 m.
 
ichivictus said:
That must have been it. I just redid the problem and received an answer a little greater than 12 meters. Not sure where I went wrong exactly, but no worries.

Thanks :)
I suggest you redo the problem again. You didn't show your work, so there's no knowing where you made a mistake.

Hint: Temporarily ignore the fact that the observer was 5.1 meters above the ground. Suppose you threw a ball upwards at 2.9 m/s. How high will it fly given g=9.82 m/s2?
 

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