Velocity/acceleration in calc via derivatives

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Homework Help Overview

The problem involves analyzing the motion of a ball thrown vertically upward, with a focus on determining the time at which it reaches maximum height and the value of that maximum height. The context is rooted in calculus, specifically the application of derivatives to understand velocity and acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the condition for maximum height, specifically referencing the derivative of height with respect to time. There is an exploration of the relationship between velocity and the height of the ball, with questions about the implications of velocity being zero.

Discussion Status

The discussion is active, with participants confirming the condition for maximum height and engaging in clarifying questions about the relationship between height and velocity. Some participants express confidence in the problem's simplicity, while others seek deeper understanding of the underlying concepts.

Contextual Notes

There is mention of a specific time when the ball hits the ground and the associated velocity, which may influence the understanding of the motion. The original poster indicates a lack of resources due to a friend's absence, which may limit their ability to engage fully with the problem.

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Homework Statement



A ball is thrown vertically upward from the roof of a building 68.6 m high, with an initial velocity of 24.5m/s. Its height above the ground at time, t=>0, is given by

h(t) = 68.6 + 24.5t-4.9t^2

b) When does the ball reach it's maximum height?
c) Find the maximum height reached by the ball.

Homework Equations





The Attempt at a Solution



Well, i lent my friend the note i had for this and he went away on vacation, therefore I'm stuck with these 2 questions.

Some notes: ball hits the ground in 7 seconds, when the velocity is -44.1m/s
v(t) = -9.8t+24.5
a(t)=-9.8
 
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The ball reaches it's maximum height when h'(t)=0. Right? Can you find t?
 
dick said:
the ball reaches it's maximum height when h'(t)=0. Right? Can you find t?

2.5time, height 99.225
 
Last edited:
Ok, that was fast. Not so bad a problem, yes?
 
Yes, it was easy, but can I ask how you knew the ball reached its maximum height when velocity was 0?
 
As long as the velocity is positive it is still going up and hasn't yet reached its highest point. When the velocity is negative is is already going down and has passed its highest point!
 
HallsofIvy said:
As long as the velocity is positive it is still going up and hasn't yet reached its highest point. When the velocity is negative is is already going down and has passed its highest point!

So basically, when h'(t) = 0 , then h(t)'s y value will be at its highest point?
 

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