• Jun 11th 2006, 09:49 AM
usm_67
a model rocket is launched from the roof of a bulinding. its flight path through air is modelled bu h=-5t^2+30t+10 where h is height of the rocket above ground in thr metres and t is the time after the launch in seconds.

(a) when will teh rocket hit the ground?
(b) when is the rocket 45m off the ground?
• Jun 11th 2006, 10:14 AM
Soroban
Hello, usm_67!

Quote:

A model rocket is launched from the roof of a building.
Its flight path through air is modelled by $h = -5t^2+30t+10$
where $h$ is height of the rocket above ground in metres and $t$ is the time after the launch in seconds.

(a) When will the rocket hit the ground?
When the rocket hits the ground, $h = 0$.

Solve: . $-5t^2 + 30t + 10 \:=\:0$

Divide by -5: . $t^2 - 6t - 2\:=\:0$

Quadratic Formula: . $t\:=\:\frac{-(-6) \pm\sqrt{(-6)^2 - 4(1)(-2)}}{2(1)}\;=\;\frac{6\pm\sqrt{44}}{2}$

Therefore: . $t\;=\;3 + \sqrt{11}\;\approx\;6.3$ seconds.

Quote:

(b) When is the rocket 45m off the ground?
When is $h = 45\;?$

Solve: . $-5t^2 + 30t + 10\;=\;45$

We have: . $-5t^2 + 30t - 35\;=\;0\quad\Rightarrow\quad t^2 - 6t + 7\;=\;0$

Quadratic Formula: . $t\;=\;\frac{-(-6) \pm\sqrt{(-6)^2 - 4(1)(7)}}{2(1)}\;=\;\frac{6 \pm\sqrt{8}}{2}$

Therefore: . $t\;=\;3 + \sqrt{3}\;\approx\;4.4$ seconds.
• Jun 11th 2006, 10:51 AM
usm_67
do we have to always divide by common factor in this case dat is 5 but we r not using 5 anywhere and one more thing if we change this to in vertex form than how can we find the solution for a and b.
• Jun 11th 2006, 11:05 AM
earboth
Quote:

Originally Posted by usm_67
do we have to always divide by common factor in this case dat is 5 but we r not using 5 anywhere and one more thing if we change this to in vertex form than how can we find the solution for a and b.

Hello,

I haven't got much time. So I will only show you step by step how you can get the vertex form of your parabol:

$h = -5t^2+30t+10$

$h=-5(t^2-6t-2)=-5(t^2-6t+9-9-2)$ = $-5((t-3)^2-11)$

$h=-5(t-3)^2+55$

So the vertex is V(3,55). That means after 3 seconds the rocket had reached it's greatest height: 55 m.

Greetings

EB
• Jun 11th 2006, 11:08 AM
usm_67
k thanks i got it
• Jun 12th 2006, 05:30 AM
topsquark
Quote:

Originally Posted by Soroban
Quadratic Formula: . $t\;=\;\frac{-(-6) \pm\sqrt{(-6)^2 - 4(1)(7)}}{2(1)}\;=\;\frac{6 \pm\sqrt{8}}{2}$

Therefore: . $t\;=\;3 + \sqrt{3}\;\approx\;4.4$ seconds.

I didn't check through all of this (as my fancy calculator is still packed and I'm too lazy to unpack it! :) ), but I see two errors.

The first is that $\frac{6 \pm\sqrt{8}}{2}$ = $\frac{6 \pm 2 \sqrt{2}}{2}$ = $3 \pm \sqrt 2$.

The second error is in using only the "+" sign. The initial height of the rocket is 10 m and the maximum height of the rocket is 55 m. Thus there will be TWO times that the rocket is at 45 m. The first is at t = $3 - \sqrt{2}$ s (on it's way up) and the second is at t = $3 + \sqrt{2}$ s (on it's way back down.)

-Dan
• Jun 12th 2006, 07:32 AM
Soroban
Hello, topsquark!

Quote:

. . . but I see two errors.
Guilty as charged!

The first is a simple typo . . . *blush*

The second is a really stupid oversight . . . (*blush*)²
• Jun 12th 2006, 02:31 PM
topsquark
Quote:

Originally Posted by Soroban
Hello, topsquark!

Guilty as charged!

The first is a simple typo . . . *blush*

The second is a really stupid oversight . . . (*blush*)²

Nothing stupid about it. I've been catching my students on that one for YEARS! :D

-Dan