High School Calculus Velocity question

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SUMMARY

The discussion centers on a high school 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. The equation used to model the ball's motion is -4.9t² - 22t + 443 = 0, leading to a time of approximately 7.5 seconds to reach the ground. The speed at impact is calculated using the first derivative, resulting in a final speed of 95.5 m/s. The solution approach was confirmed as correct, with a clarification on the use of derivatives.

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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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