High School Calculus Velocity question

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Homework Help Overview

The discussion revolves around a calculus problem involving the motion of a ball thrown straight down from a height of 443 meters with an initial velocity of 22 m/s. Participants are analyzing the time it takes for the ball to reach the ground and its speed upon impact, utilizing equations related to gravitational motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss setting up the equation -4.9t^2 - 22t + 443 = 0 to find the time of impact and the speed at impact using derivatives. There is a focus on verifying the correctness of the approach and the interpretation of derivatives.

Discussion Status

Some participants express agreement with the original poster's approach, while others question the classification of the derivative used in the calculations. This indicates an ongoing exploration of the problem's setup and the mathematical reasoning involved.

Contextual Notes

There is a repeated emphasis on the same problem statement and equations, suggesting a need for clarity in understanding the derivatives and their applications in the context of the motion described.

Noriko Kamachi
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Homework Statement


If you throw a ball straight down from a building that is 443 meters tall with a velocity of 22 m/s how long would it take to reach the ground, and what would be its speed at impact?

Homework Equations


Since the ball is being thrown down with a velocity of 22 m/s I plug that into the gravitational constant equation for Earth -4.9t^2 -22t +443.

The Attempt at a Solution


To get when the ball hits the ground I set -4.9t^2 -22t +443=0. Solving for t I get about ~7.5. To get the speed at impact I took the second derivative which is -9.8t -22. Plugging in t into this equation I get |-9.8(7.5) -22| for 95.5 m/s for the speed at impact. Was this problem handled in the correct fashion?
 
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Noriko Kamachi said:

Homework Statement


If you throw a ball straight down from a building that is 443 meters tall with a velocity of 22 m/s how long would it take to reach the ground, and what would be its speed at impact?

Homework Equations


Since the ball is being thrown down with a velocity of 22 m/s I plug that into the gravitational constant equation for Earth -4.9t^2 -22t +443.

The Attempt at a Solution


To get when the ball hits the ground I set -4.9t^2 -22t +443=0. Solving for t I get about ~7.5. To get the speed at impact I took the second derivative which is -9.8t -22. Plugging in t into this equation I get |-9.8(7.5) -22| for 95.5 m/s for the speed at impact. Was this problem handled in the correct fashion?

It looks good to me.
 
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Noriko Kamachi said:

Homework Statement


If you throw a ball straight down from a building that is 443 meters tall with a velocity of 22 m/s how long would it take to reach the ground, and what would be its speed at impact?

Homework Equations


Since the ball is being thrown down with a velocity of 22 m/s I plug that into the gravitational constant equation for Earth -4.9t^2 -22t +443.

The Attempt at a Solution


To get when the ball hits the ground I set -4.9t^2 -22t +443=0. Solving for t I get about ~7.5. To get the speed at impact I took the second derivative which is -9.8t -22. Plugging in t into this equation I get |-9.8(7.5) -22| for 95.5 m/s for the speed at impact. Was this problem handled in the correct fashion?
-9.8t-22 is not the second derivative; it's the first derivative of the position function, -4.9t2-22t+443 = 0.
 
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Dick said:
It looks good to me.

Thanks! :D

SteamKing said:
-9.8t-22 is not the second derivative; it's the first derivative of the position function, -4.9t2-22t+443 = 0.

Ah I see! Thanks for correcting me!
 

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