Changing subject of equation: h(t) = -4.9t2 + 30t

A

Aaron.f

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So im 24 and just starting a calculus course as a bring home course.
Its been a while since ive done any of this so i apologize if this is
basic knowledge:p

I have an equation to find the height of a ball thrown with the given time
h(t) = -4.9t2 + 30t

the one question is when does the ball hit the ground

what is the proper way to change the subject of that formula to find height
 
Aaron.f said:
I have an equation to find the height of a ball thrown with the given time
h(t) = -4.9t2 + 30t

the one question is when does the ball hit the ground

what is the proper way to change the subject of that formula to find height
Click to expand...
The equation already is "solved in terms of height"; namely, the "h(t)", being the height at time t.

This question is basic algebra. It sounds as though you have forgotten what function notation is, how to evaluate, and how to convert between English and math. For instance, when the question asks you "when" something happens, it is asking for the time, not the height.

To review this specific topic (being projectile-motion quadratic word problems), try here. Then, noting what the height must be for the indicated time, solve the quadratic for the requested value.

If you get stuck, please reply showing all of your steps so far. Thank you! ;)
 
Oh wow oops, just been a stressfull day i had a brain fart:p i do understand that the original equation is to find height already:p. Sorry what i meant to ask really was that they want me to find what the time is when height is 0.
this take home course book has certain questions that i dont hand in and the answers are in the back.
with how much work it showed in the answer i couldnt quite see how they change the subject to find time.
Mostly im just stuck on the idea of what you do with the two ts being on the same side but one having an exponent
 
If height = 0 you need to solve the equation \(\displaystyle -4.9t^2 +30t=0\).

Now factorise the LHS. Do you know how to do that?

Then apply the NULL FACTOR LAW which says: "If ab=0 then either a=0 or b=0."

You will get 2 solutions for t, and you need to interpret them in the context of the question.
 
Thank you so much. I feel kinda silly i know i seem a little out of my league here
its been 8 years or so since ive seen any math so im hoping itl all start coming back to me.
gotta find a way though its what i need for my career, so i really appreciate the help:)
 
Aaron.f said:
Thank you so much. I feel kinda silly i know i seem a little out of my league here
its been 8 years or so since ive seen any math so im hoping itl all start coming back to me.
gotta find a way though its what i need for my career, so i really appreciate the help:)
Click to expand...

I always tell my students that there is no such thing as a silly question, except for the question that you don't ask!
Did you complete this problem?
 
Actually i did not :(. I do remember factoring equations but ive definitely forgotten some of it.
I went to try it out after your first reply. But its hasnt worked out im getting something wrong.
How would the LHS be factored?
 
Well i really havent tried much for the first while i thought i needed
to change the subject. Now that i realize that wasnt right i attempted to
factor the LHS but as far as I thought and could find online those all look like
ax2 + bx + c. So for this equation what would c be. I have no work i can show
because i cant get passed not having anything for c
 
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Well, okay, so let's think about what's going on for a second and see if we can't figure it out. You started with -4.9t2+30t and you want to transform it into a form like at2+bt+c. You've correctly identified that we'd have a=-4.9 and b=30. But what is c? Hm... well, what if we just leave it as a symbol for now?

-4.9t2+30t+c

Let's think about what value c might have. We can't add anything because that would change the value of the expression, so c can't be a positive number. We can't subtract anything either because that would also change the value of the expression, so c can't be a negative number. That means what we need to do is add... nothing. Is there some value c can take on that would be the same as adding nothing?

Now, that all being said, you don't actually need to transform this expression into ax2+bx+c to factor it. You can, but in this case it only adds unnecessary work and complication for you. Instead, let's see if we can find an easier way. If you had been given the expression y(2y+4) you could multiply that out and arrive at 2y2+4y. Now suppose you'd instead been given it in the form 2y2+4y, and you wanted to run the process "in reverse." You'd look and see that there's a common element of both terms (a 'y') and "pull out" that commonality, arriving back at y(2y+4). Now, how do you think this relates to the problem you were actually given? Is there a common element of both terms?
 
Aaron.f said:
...online those all look like ax2 + bx + c. So for this equation what would c be...
Click to expand...

-4.9t^2 + 30t + 0

c = 0

But, we don't need to use a factoring method for trinomials; we can ignore that zero, and simply factor the given binomial.

-4.9t^2 + 30t

We need to examine the factors in each term, to look for "common factors". Common factors are factors that appear in each term. Let's write out the factors in each term.

(-1)(4.9)(t)(t) + (30)(t)

Looking at the factors above, we see that (t) is the only factor that appears in each term. So, the "common factor" is t, and we factor it out.

(t)(-4.9t + 30)

If you google "factoring binomials", you will find videos and lessons with a lot of examples to study.

Now, Harry the Cat told you about the Null Factor Law. In the USA, we call it the Zero-Product Property. It tells us that, if the product of two quantities is zero, then at least one (or possibly both) quantities must be zero.

We have (t)(-4.9t + 30) = 0

Therefore:

t = 0

or

-4.9t + 30 = 0

The first equation t=0 tells us that the ball is at ground level at time zero, which is somewhat peculiar, but maybe the person tossing the ball is standing in a big hole.

The second equation -4.9t + 30 = 0 is what you need to solve, to find how much time it takes for the ball to return to ground level.

I strongly encourage you to review factoring binomials, before moving on to other exercises. :cool:
 
Ok so that does make sense now and im going to study that like youve suggested
so im sure i have a fuller understanding of all the concepts going forward

now when solving -4.9t +30 = 0 i would isolate t
-4.9t +30 =0
-4.9t = 0 -30
t = -30/4.9

it should be positive 30/4.9 what have i done wrong there

thank you everyone for youre help. I really appreciate youre time
 
Aaron.f said:
Ok so that does make sense now and im going to study that like youve suggested
so im sure i have a fuller understanding of all the concepts going forward

now when solving -4.9t +30 = 0 i would isolate t
-4.9t +30 =0
-4.9t = 0 -30
t = -30/4.9 .... You need to divide both sides by -4.9 in this step. This will give t = -30/-4.9 = 30/4.9

it should be positive 30/4.9 what have i done wrong there

thank you everyone for youre help. I really appreciate youre time
Click to expand...
see comment in red
 
If you check out the control panel or engines in your airplane called Calculus, you'll see a lot of algebra. If you've forgotten your algebra, it's going to be difficult to get off the ground. Complicating matters is that your at-home course comes with no co-pilot; you have only the control tower to guide you, and they don't have time to explain algebra.

:idea: We're here to help, but reviewing algebra first is best.
 
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