Velocity/acceleration in calc via derivatives

In summary, the conversation discusses finding the maximum height and time at which a ball thrown vertically from a building reaches its highest point. The ball's height and velocity equations are given, and it is determined that the ball reaches its maximum height when its velocity is 0. The maximum height is found to be 99.225m and the time at which it reaches this height is 2.5 seconds. The concept of velocity and its relation to the ball's height is also discussed.
  • #1
Draggu
102
0

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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  • #2
The ball reaches it's maximum height when h'(t)=0. Right? Can you find t?
 
  • #3
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:
  • #4
Ok, that was fast. Not so bad a problem, yes?
 
  • #5
Yes, it was easy, but can I ask how you knew the ball reached its maximum height when velocity was 0?
 
  • #6
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!
 
  • #7
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?
 

1. What is the difference between velocity and acceleration?

Velocity is the rate of change of an object's displacement over time, while acceleration is the rate of change of an object's velocity over time. In other words, velocity measures how fast an object is moving and in what direction, while acceleration measures how quickly an object's velocity is changing.

2. How are velocity and acceleration related to derivatives?

In calculus, velocity is the derivative of position with respect to time, and acceleration is the derivative of velocity with respect to time. This means that velocity and acceleration can be calculated using the principles of differentiation, which involve finding the slope of a curve at a given point.

3. What is the significance of using derivatives in calculating velocity and acceleration?

Using derivatives allows us to find the instantaneous velocity and acceleration of an object at a specific moment in time, rather than just its average velocity and acceleration over a longer period. This is important in understanding the motion of objects that are constantly changing their speed and direction.

4. How do we use derivatives to find the velocity and acceleration of an object?

To find the velocity of an object, we take the derivative of its position function with respect to time. Similarly, to find the acceleration of an object, we take the derivative of its velocity function with respect to time. This gives us the instantaneous values of velocity and acceleration at a given time.

5. Can derivatives be used to analyze non-uniform motion?

Yes, derivatives can be used to analyze the velocity and acceleration of objects that are not moving at a constant speed or in a straight line. By taking the derivatives of the position function, we can still find the instantaneous velocity and acceleration of the object at any given time, even if its motion is not uniform.

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