Thread: Word problems

A model rocket is launched from the roof of a building. It's flight path through the air is modeled by h=-5t²+23t+10, where h is the height of the rocket above ground in meters and t is the time after the launch in seconds.

A) when does the rocket hit the ground?

I got this far H= -1(5t²-23t-10) -23-50
H= -1(t-25)(t+2) -25+2
The teacher answer has this..........
H= -1(5t²-23t-10)
H= -(t-5)(5t+2)

t-5= 0 5t+2=0
t=5 t=-2/5

Where does 5t come from?
I don't get the whole H= -5t²+23t+10 equation what does -5 stand for? 23t? 10?
Teacher just taught me the equation and I follow it but, I kinda dont get the whole thing.

How did you get these steps?

Originally Posted by buck

I got this far H= -1(5t²-23t-10) -23-50
H= -1(t-25)(t+2) -25+2

All you need to do is solve where $\displaystyle h=0 $

i.e. $\displaystyle 0=-5t²+23t+10$

You can factor the RHS and apply the null factor law like in your teachers solution or just use the quadratic formula.

Which method do you prefer?

I don't know my teachers method. I don't know which method I use but if I bring down t-25=0 wouldn't 5 be 25?
My teacher got a totally different answer because of the (5t+2)
i dont know how she got that

Originally Posted by buck

I don't know which method I use but if I bring down t-25=0 wouldn't 5 be 25?

you may have to clarify further what you mean by this.

I would solve using the quadratic formula $\displaystyle \displaystyle t=\frac{-23\pm\sqrt{23^2-4\times -5\times 10}}{2\times -5}$