Velocity/acceleration in calc via derivatives

[Draggu]

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

 

[Dick]

The ball reaches it's maximum height when h'(t)=0. Right? Can you find t?

 

[Draggu]

the ball reaches it's maximum height when h'(t)=0. Right? Can you find t?

2.5time, height 99.225

 

[Dick]

Ok, that was fast. Not so bad a problem, yes?

 

[Draggu]

Yes, it was easy, but can I ask how you knew the ball reached its maximum height when velocity was 0?

 

[HallsofIvy]

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!

 

[Draggu]

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?