When will the ball hit the ground?

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homeschool girl

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The equation
$y = -6t^2 - 10t + 56$
describes the height (in feet) of a ball thrown downward at 10 feet per second from a height of 56 feet from the surface from Mars. In how many seconds will the ball hit the ground? Express your answer as a decimal rounded to the nearest hundredth.

I don't know how to start this problem, I tried 5.6 but that was wrong, and I don't know how the -6t^2 comes into play.
 
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I don't know how to start this problem, I tried 5.6 but that was wrong, and I don't know how the -6t^2 comes into play.
 
Where does the problem come from? What are you studying? Are there examples in the textbook? Why do you think this involves "direct proportion"?
 
What must \(y\) be when the ball is on the ground?
 
lev888 said:
Where does the problem come from? What are you studying? Are there examples in the textbook? Why do you think this involves "direct proportion"?
Click to expand...
sorry it didn't say what the question was but there was another question next to it with the title of direct proportions and I made a mistake bc I couldn't find my glasses
 
homeschool girl said:
sorry it didn't say what the question was but there was another question next to it with the title of direct proportions and I made a mistake bc I couldn't find my glasses
Click to expand...

I've edited the thread title to reflect the nature of the question being asked.

homeschool girl said:
zero ?
Click to expand...

Correct, so we then have:

[MATH]-6t^2-10t+56=0[/MATH]
I would choose to divide through by -2 to get:

[MATH]3t^2+5t-28=0[/MATH]
Have you been taught how to solve quadratics?
 
Let's factor:

[MATH](3t-7)(t+4)=0[/MATH]
Yes, I agree with the roots you found. What should we do with the negative root?
 
You should always ask yourself what do the variables mean. Obviously t = time in seconds, But you may have to think a moment to define y. It is the height of the ball above ground in feet. Once you have determined what that implicitly defined variable represents and jotted it down so you do not forget it, it quickly becomes obvious that when the ball touches the ground, y = 0.

That is how Mark came up with

[MATH]-\ 6t^2 - 10t + 56 = 0.[/MATH]
 
Well, yes, but I mean look at the question, at \(t=0\), we have \(y=56\). So, this is at the moment the ball is thrown. So, \(t=-4\) is 4 seconds before the ball is thrown. We are to assume that the equation relating the height of the ball above the ground to the elapsed time after the ball is thrown does not apply to any time before the ball is thrown, only at the time it is thrown until the moment it strikes the ground.

There is a physical interpretation to the negative root, but you don't need to worry about that right now. For now, you simply need to discard any negative roots. You will need to keep in mind when solving problems what the variables represent. You will find, for example, when working with physical measurements of distance, that negative roots will need to be discarded, because a negative measurement of distance is usually meaningless.
 
homeschool girl said:
So we just have t= 7/3 ?
Click to expand...
Correct.

Since the other choice (t = -4) is invalid - the only choice left to us is t = 7/3

To complete the problem, I would evaluate the function (y = − 6*t2−10*t + 56) at t = 7/3 and confirm that y is indeed equal to 'zero' at that time.
 
As already pointed out you need to understand what the equation is telling you. The left hand side is y which is given to be the number of feet above the ground. The right hand side is something in terms of t. t was given to be time in seconds. Now if you plug in some value for t and crank out the right hand side you get some value. But y equals whatever the right hand side equal. That is after plugging in some value for t on the rhs you get the value for y WHICH IS the height above the ground for that t value.

Now you want to know when the ball hits the ground. Since you want to know when, then you want to know what value t should be. Now when the ball hits the ground y must be 0. So you need to solve 0 = − 6*t^2−10*t + 56.
As far as understanding where the -6t^2 came from you do not need to know that answer. That equation was given to you and you just need to work with it. This negative parabola will reach 0 twice. You need to realize that if you dropped a ball from a height that it will only hit the ground once (we are NOT considering the ball bouncing up and then hitting the ground again and again and again..). So if you get two answer then one must be rejected. You dropped the ball at t=0 seconds so how can the ball hit the ground at t=-4 seconds? The experiment hasn't even started at t=-4!!

I hope this is all clear to you. If not then please post back stating where you are confused.
 
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