# ANOTHER parabola related question! ._.

1. Oct 1, 2009

### Stripe

1. The problem statement, all variables and given/known data
a cannon launches a projectile. H, is the height in metres that the cannon ball is above the horizon by t seconds.

h= -0.75t^2 + 16t + 3

2. Relevant equations
What is the maximum height reached by the cannonball?
3. The attempt at a solution
Ok so i tried completing the square:
h= -(0.75t^2-16t-3)
h= -(0.75^2-16t+64)+64+3 < (Is this right, that instead of -64 it is + because in the first brackets i was -64 really?)
h= -(0.75t-8)^2+67

Therefore the turning point is (8,67)?

So would the highest point not be 67 metres? because when i graph it in my calculator it is 88.333 or something like that.

Thanks again >_<

2. Oct 2, 2009

### Staff: Mentor

What you want to do is to factor out the coefficient of your squared term and the first-degree term, like so:
h= -(3/4)t2 + 16t + 3
= -(3/4)[t2 - (64/3)t + ___] + 3
= -(3/4)[t2 - (64/3)t + (322/9)] + 3 + (3/4)(322/9)
and so on.

3. Oct 2, 2009

### Office_Shredder

Staff Emeritus
Square (.75t-8). You don't get what you think you get. Try completing the square again, and keep in mind that the coefficient of t2 is not 1 here! You need to adjust for that

4. Oct 2, 2009

### Stripe

oh shoop da woop thank you guys so much!

After deciphering some of your maths jingo, i got it! Thanks guys i really appreciate your help!

and Mark a special thanks to you for replying to all my threads so hastily!

Thanks guys :D