1. ## Trig

A ball is thrown verically upward from the balcony of an apartment building. The ball falls to the ground the height of the ball, h metres, above the ground after t seconds is given by the function: h(t)= -5t^2+15t+45

I have the factored form or vertex cant remeber which one it is: y=-5(x+2.3)+56.5

I'm having trouble finding how long it takes the ball to reach the max height. and how high the balcony is
any help is appreciated thnks

2. Originally Posted by Gomaz23
A ball is thrown verically upward from the balcony of an apartment building. The ball falls to the ground the height of the ball, h metres, above the ground after t seconds is given by the function: h(t)= -5t^2+15t+45

I have the factored form or vertex cant remeber which one it is: y=-5(x+2.3)+56.5

I'm having trouble finding how long it takes the ball to reach the max height. and how high the balcony is
any help is appreciated thnks
Your vertex form is wrong...How are you getting that?

3. really? well i go tit by factoring the orginal function.

4. Originally Posted by Gomaz23
really? well i go tit by factoring the orginal function.
...By completing the square you mean? show me all your steps and then we can work at it

5. h(t)= -5t^2+15t+45

next: -5(t^2-3t) (45
all i did was divide out the 5 and divide the 15 by negative 5 and took out the 45
next: -3/2=-1.5 -1.5^2= 2.3
divide negative 3 by 2 then squared the answer
next: -5(t^2-3t+2.3) -2.3 (-5) (45
-5(t^2-3t+2.3) 11.5 (45
i took out the 2.3 then multiplied the negative 5 to it
Next: -5(t^2-3t+2.3) 56.5
next: y= -5(x+2.3) +56.5

6. Originally Posted by Gomaz23
h(t)= -5t^2+15t+45

next: -5(t^2-3t) (45
all i did was divide out the 5 and divide the 15 by negative 5 and took out the 45
next: -3/2=-1.5 -1.5^2= 2.3
divide negative 3 by 2 then squared the answer
next: -5(t^2-3t+2.3) -2.3 (-5) (45
-5(t^2-3t+2.3) 11.5 (45
i took out the 2.3 then multiplied the negative 5 to it
Next: -5(t^2-3t+2.3) 56.5
next: y= -5(x+2.3) +56.5
close

$\displaystyle h(t)-5t^2+15t+45$

$\displaystyle =-5(t^2-3t)+45$

$\displaystyle =-5(t^2-3t+2.25-2.25)+45$

$\displaystyle =-5(t^2-3t)+45+11.25$

$\displaystyle =-5(t-1.5)^2+56.25$

$\displaystyle V1.5,56.25)$

7. now how do i find how long it takes to reach the max value and how high the balcony is

8. Originally Posted by Gomaz23
now how do i find how long it takes to reach the max value and how high the balcony is
I) 1.5 seconds, the x co-ordinate of the vertex

II)The ball leaves the balcony at $\displaystyle t=0$

Thus
h=-5tsqrd+15+45
h=-5(0)sqrd+15+45
=45m

Therefore the balcony is 45m high

9. would i need to plug this in to find the time the ball reached that height

10. Originally Posted by Gomaz23
would i need to plug this in to find the time the ball reached that height
I modified my post, it is 1.5seconds, the y-coordinate, i misread the question at first

11. o ok thnks a lot

12. Originally Posted by Gomaz23
o ok thnks a lot
No problem, have a good one

13. wait one more question howd u get 1.5 for the vertex form

14. Originally Posted by Gomaz23
wait one more question howd u get 1.5 for the vertex form
when you get to the end which was

-5(tsqrd-3)+56.25

you must divide 3 by 2 to yield

-5(t-1.5)sqrd+56.25

you will always divide k by 2 for the final in y=a(h-k)+d

just a rule of completing the square