High School Calculus Velocity question

In summary, to calculate the time it takes for a ball thrown straight down from a 443 meter tall building with a velocity of 22 m/s to reach the ground, the equation -4.9t^2 -22t +443=0 can be used by setting it equal to 0 and solving for t. The speed at impact can be calculated by taking the first derivative of the position function, -4.9t^2 -22t +443, and plugging in the value of t found earlier. The correct speed at impact in this scenario is 95.5 m/s.
  • #1
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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  • #2
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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  • #3
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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  • #4
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!
 

FAQ: High School Calculus Velocity question

1) What is velocity in high school calculus?

Velocity in high school calculus is a measure of the rate at which an object's position changes over time. It is a vector quantity that includes both the speed and direction of an object's motion.

2) How is velocity calculated in high school calculus?

In high school calculus, velocity is calculated by taking the derivative of an object's position function with respect to time. This means finding the slope of the tangent line to the position curve at a specific time.

3) What is the difference between average velocity and instantaneous velocity?

Average velocity is the total displacement of an object divided by the total time it took to move that distance. Instantaneous velocity, on the other hand, is the velocity at a specific moment in time and is found by taking the derivative of the position function.

4) How is velocity related to acceleration in high school calculus?

Velocity and acceleration are related by the fundamental theorem of calculus. Acceleration is the rate of change of velocity over time, and velocity is the rate of change of position over time. Therefore, acceleration is the second derivative of the position function.

5) How can velocity be represented graphically in high school calculus?

Velocity can be represented as the slope of the tangent line to a position curve on a graph. It can also be represented by a vector arrow, where the length of the arrow represents the speed and the direction of the arrow represents the direction of motion.

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