# Math Help - finding height reached by cannon shell

1. ## finding height reached by cannon shell

a cannon fires a shell that rise (h) cm after t seconds where
$h(t)=800t-16t^2$

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

2. 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.

3. am tryin to determine the constants A and B but am having no luck

4. 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 $-16(t-25)^2+1000$

so i can assume that the maximium point is (25, 1000) is that correct?

5. 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 $-16(t-25)^2+1000$

so i can assume that the maximium point is (25, 1000) is that correct?

6. $16*625= 100^2= 10000$, not 1000.

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

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

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

$0 = 16t(50-t)$

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

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

$h(25) = 16(25)[50-25] = 16 \cdot 25^2 = 10000$

8. 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?

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

how would i find the maximium height (k) reached by the shell and how would its graph look. for $h=800t -16t^2$
$x = ut+\frac{1}{2}at^2$

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

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

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

so $\rightarrow 800-32t=0$

rearrange to find t which reaches max height.

which comes to 25s.

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

which comes to 10000m.

10. 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.

11. 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)