finding height reached by cannon shell

• Apr 9th 2010, 10:17 PM
sigma1
finding height reached by cannon shell
a cannon fires a shell that rise (h) cm after t seconds where
$\displaystyle h(t)=800t-16t^2$

how would i find the maximium height (k) reached by the shell and how would its graph look. for $\displaystyle h=800t -16t^2$
• Apr 10th 2010, 02:53 AM
HallsofIvy
You should be able to see that the graph is a parabola.

Complete the square so you can write it as h= -16(t- a)^2+ b for some a and b. Since a square is never negative, h is always b minus something. h will be largest when t= a.
• Apr 10th 2010, 05:20 AM
sigma1
am tryin to determine the constants A and B but am having no luck
• Apr 10th 2010, 06:00 AM
sigma1
Quote:

Originally Posted by HallsofIvy
You should be able to see that the graph is a parabola.

Complete the square so you can write it as h= -16(t- a)^2+ b for some a and b. Since a square is never negative, h is always b minus something. h will be largest when t= a.

i got $\displaystyle -16(t-25)^2+1000$

so i can assume that the maximium point is (25, 1000) is that correct?
• Apr 10th 2010, 06:12 AM
sigma1
Quote:

Originally Posted by HallsofIvy
You should be able to see that the graph is a parabola.

Complete the square so you can write it as h= -16(t- a)^2+ b for some a and b. Since a square is never negative, h is always b minus something. h will be largest when t= a.

i got $\displaystyle -16(t-25)^2+1000$

so i can assume that the maximium point is (25, 1000) is that correct?
• Apr 11th 2010, 03:59 AM
HallsofIvy
$\displaystyle 16*625= 100^2= 10000$, not 1000.
• Apr 11th 2010, 05:47 AM
skeeter
Quote:

Originally Posted by sigma1
a cannon fires a shell that rise (h) cm after t seconds where
$\displaystyle h(t)=800t-16t^2$

how would i find the maximium height (k) reached by the shell and how would its graph look. for $\displaystyle h=800t -16t^2$

$\displaystyle h(t) = 800t - 16t^2$

$\displaystyle h(t) = 16t(50 - t)$

$\displaystyle 0 = 16t(50-t)$

note that the $\displaystyle h = 0$ at $\displaystyle t = 0$ and $\displaystyle t = 50$

due to the parabola's symmetry, the shell will be at its highest point at $\displaystyle t = 25$, midtime between the zeros.

$\displaystyle h(25) = 16(25)[50-25] = 16 \cdot 25^2 = 10000$
• Apr 11th 2010, 06:09 AM
sigma1
ooh thank you i now see that.. so i can say then that the maximium height reached is 10000 as opposed to stating the coordinates of (25,10000),

i have also sketched this graph is it correct?
• Apr 11th 2010, 09:01 AM
BabyMilo
Quote:

Originally Posted by sigma1
a cannon fires a shell that rise (h) cm after t seconds where
$\displaystyle h(t)=800t-16t^2$

how would i find the maximium height (k) reached by the shell and how would its graph look. for $\displaystyle h=800t -16t^2$

$\displaystyle x = ut+\frac{1}{2}at^2$

$\displaystyle \frac{dx}{dt}= v = u+at$

so we have $\displaystyle x=800t-16t^2$
$\displaystyle \frac{dx}{dt}= v = 800-32t$

when $\displaystyle v=0$ it reaches max. height because when v is negative, it will start falling.

so $\displaystyle \rightarrow 800-32t=0$

rearrange to find t which reaches max height.

which comes to 25s.

then sub it into $\displaystyle x=800t-16t^2$, to find max height.

which comes to 10000m.
• Apr 11th 2010, 09:14 AM
HallsofIvy
Quote:

Originally Posted by sigma1
ooh thank you i now see that.. so i can say then that the maximium height reached is 10000 as opposed to stating the coordinates of (25,10000),

i have also sketched this graph is it correct?

Yes, that is correct.
• Apr 11th 2010, 09:46 AM
sigma1
Quote:

Originally Posted by HallsofIvy
Yes, that is correct.

thanks for that. but how would these transformation make the graph. look .could you show me?

|h(t)|

h(t+30)