Quadratic Equation?

Question

A rose thrown onto a platform has a height, h(t), given by the formula -16t^2 + 32t + 20. How long will it take before the rose hits the ground?

What I did with past problems on the worksheet was this:
h(t)=18x-6x^2
h(t)=-6x (-3 + x)
x=0, 3
3 seconds

But it doesn't seem to work with this problem, so I decided to ask here.

PLEASE, don't just post the answer, show the work, explaining a bit if needed. Or, you don't even have to post the answer, just tell me how you go about doing this problem.

Thanks ^_^

I meant from the platform, sorry. Thanks for the answers, I thought it had to be like the previous problems (see example) and that got me confused.

I do understand the quadratic formula, I just didn't know that it could be used in this situation. I wish I could give out multiple best answers, but oh well...you guys helped me a lot ^_^

Answer 1

When the rose hits the ground the height would be 0...so h(t) = 0.

If h(t) = 0...and the formula for h(t) is -16t^2 + 32t + 20...

then 0 = -16t^2 + 32t + 20

Solve...

Now that I read it a second time...the problem says a rose is thrown onto a platform. How would it hit the ground if it's on the platform? (assuming it's not moved)

Answer 2

You have a slight problem with your original solution. You know, the time in second is expressed in "t".

The formula is h(t) but you are looking for the function of h(t) becomes zero, so you will be solving

0 = -16t^2 + 32t + 20, more commonly written as:
-16t^2 + 32t + 20 = 0

You should eventually get to:
x= something (you said to not give you the answer)

One of the answer will be a negative answer. You should reason that this is not a valid answer because it means it happened in the past. (negative time)

The other answer will be a positive answer. It will be the figure you are looking for.

By the way, use the quadratic equation to arrive at two X values.

Answer 3

You can either use the quadratic equation and you can find that formula online. But what I do with those problems is multiply the first and last term together. Find the factors of that term that will add together to give you the middle term
For example
2x^2 + 9x + 10.
2 * +10 = 20. Factors of 20 are
1, 20
2, 10
4, 5

Since 4 and 5 will multiply together and get 20 and will add together to get 9, just use 4 and 5 x in place of 9x.

2x^2 + 4x + 5x + 10
Factor by grouping
2x (x + 2) + 5 (x + 2)
Factor out the gcf (x + 2)
then solve like you did the other equations.

Since you have signs, be sure to pay attention to those! :)
Hope that helps!

Answer 4

this is what i did:

h(t)= -16t^2+32t+20
I got the equation directly from the word problem

h(t)= (-4)(4t^2 - 8t - 5)
then i factored out the greatest common factor

*then used the quadratic equation and got

t= (-0.5, 2.5)

that means 2.5 seconds to hit the ground

Hope that helps

~Storm G
Kona, Hawaii

Answer 5

s = ut + (1/2)*a*t^2
Where s = distance, u = initial velocity (0 in this case), t=time, a=acceleration (g in this case = 9.8)

Now s = h = 16t^2 +32t +20

So 16t^2 + 32t +20 = (1/2)*9.8*t^2 = 4.9t^2.

So 11.1 t^2 + 32t + 20 = 0

And t is positive.

So t = -b +/- (sqrt(b^2 - 4ac)/2a)
Where b = 32, a = 11.1, c = 20

Take it from there.

Answer 6

First get rid of the - sign on the 16 by mult everything by -1 (nu don't have to but I like to do it this way)...then By using the equ'n x= -b +- sq. root (b^2-4ac) all divided by (2a) you get this as the start:
x = 32 + (then you do -) sq. root of {(-32)^2 - 4 (16) (-20)} all divided by 2(16)
I dunno if written like this you will understand but I dunno how to type up equ's...look at how the equ'n goes in ur notes or book...it's the quadratic formula. To check ur answers go here: http://www.hpcsoft.com/products/MathSoL/nonlinear/quadratic01.html