ANOTHER parabola related question ._.

[Stripe]

Homework Statement

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

Homework Equations

What is the maximum height reached by the cannonball?

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 >_<

 

[Mark44]

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.

 

[Office_Shredder]

Homework Statement

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

Homework Equations

What is the maximum height reached by the cannonball?

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 >_<

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

 

[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