# Calculus- how to find half of max height

1. Oct 23, 2011

### jtt

1. The problem statement, all variables and given/known data
the question says " when did the rock reach half its maximum height?

2. Relevant equations
a rock is thrown vertically upward from the surface of the moon at a velocity of 24m/sec and
reaches height of s= 24t-0.8t^2

3. The attempt at a solution
i figured out when the rock reaches max point which is 15 seconds. i think when velocity = zero is when you can figure out the max height. i don't know about half of max height.

2. Oct 23, 2011

### HallsofIvy

Staff Emeritus
If you know when the rock reached its maximum height, you must know what that maximum height is. Okay, divide that maximum height by 2, set your formula for height equal to that and solve. There will be two answer to that quadratic equation- the time the rock passes that height on the way up and the time it passes it on the way down.

3. Oct 23, 2011

### jtt

the time it takes to reach max height is 15 seconds and the height is 180 ft. based on how i tried to interpret the last reply this is what i got:
24-.8t^2=90
-90 -90
24t-.8t^2-90=0
is this correct?

4. Oct 23, 2011

### dynamicsolo

Yes, you'll have that quadratic equation. You'll obtain two solutions for the time, the second one being the time after the rock is thrown when it is at half the maximum height and on the way down. (As a check, the two solutions for the time should be symmetrical about the time at which the rock is at maximum height.)

5. Oct 23, 2011

### jtt

is the equation 24t-.8t^2-90=0 factorable or will i have to use the quadratic equation? i started out by dividing the whole equation by 2 and i got:
12t-.04t^2-45=0

i'm stuck and not sure where to go from here...

6. Oct 23, 2011

### dynamicsolo

It won't factor nicely: the solutions are irrational numbers. So you will need the quadratic equation.