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

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A ball is thrown upwards and passes a window at 5.1 meters with a speed of 2.9 m/s. To find the maximum height, the initial velocity (Vo) is calculated using the kinematic equation, resulting in approximately 10.01 m/s. The maximum height is then determined by considering the ball's upward motion from the window, leading to a total height of slightly over 12 meters. The discussion emphasizes the importance of correctly applying kinematic equations and suggests re-evaluating calculations for accuracy. The final approach involves adding the height gained from the window to the initial height of 5.1 meters.
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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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