s=negative t^2 plut 5t plus 500, and Jery climbed mountain

jamekadiva

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I have this problem that I can't get.

If s=negative t^2 plus 5t plus 500, and Jery has climbed a mountain that is 500ft waiten on Jane. He throws rocks off the side waiten for her to come.How long does it take for the rock to hit the water ? If 15 sec have gone by what is the height of the rock from the water? what is the highest point the rock will go by? s= distance in feet and t= time in seconds.

s= -t^2 plus 5t plus 500
 

wjm11

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Re: word problems

I have this problem that I can't get. If s=negative t^2 plus 5t plus 500, and Jery has climbed a mountain that is 500ft waiten on Jane. He throws rocks off the side waiten for her to come.How long does it take for the rock to hit the water ? If 15 sec have gone by what is the height of the rock from the water? what is the highest point the rock will go by? s= distance in feet and t= time in seconds.

s= -t^2 plus 5t plus 500
If we are working in units of feet and seconds, The general equation should be

s = -16t^2 + (v1)t + s1

where v1 is the initial velocity (for example, when we throw the rock instead of just dropping it), and s1 is the initial height (in this case 500 feet). Are you sure you have stated the problem exactly as it appears in your book?

To answer the first question, you must recognize that the height of the rock when it reaches the water is zero, i.e., s = 0. Put 0 in for s, then solve for t.

To answer the second question, put in 15 for t and find out what s equals.

To answer the third question, you must recognize that this equation will graph as a parabola that opens downward. The vertex of the parabola is where the rock is at its peak height. In other words, you must find the vertex of the parabola.

Please respond, showing all your work.
 

fasteddie65

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Re: word problems

jamekadiva said:
I have this problem that I can't get. If s=negative t^2 plus 5t plus 500, and Jery has climbed a mountain that is 500ft waiten on Jane. He throws rocks off the side waiten for her to come.How long does it take for the rock to hit the water ? If 15 sec have gone by what is the height of the rock from the water? what is the highest point the rock will go by? s= distance in feet and t= time in seconds.

s= -t^2 plus 5t plus 500
How long does it take for the rock to hit the water ?
Set s = 0 and solve for t.

If 15 sec have gone by what is the height of the rock from the water?
Set t = 15 and find s.

what is the highest point the rock will go by?
The curve s = -t^2 + 5t + 500 is a parabola. The highest point is at the vertex. The x-coordinate of the vertex can be found by using t = -b./(2a).
 

jamekadiva

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The foluma we have to use is s= negative t^2 plus 5t plus 500
he is tossing the rocks (sorry)
You sail to set s for 0 and solve for t
Is that 0 =-t^2 plus 5t plus 500
S is for distance
T is for time
(A) 0= -t^2 + 5(T) + 500

how do I find out where to put everything.I don't have a example to o by help!!
 

jamekadiva

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We have not been over the vertex I'M LOST!!
 

Subhotosh Khan

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jamekadiva said:
The foluma we have to use is s= negative t^2 plus 5t plus 500
he is tossing the rocks (sorry)
You sail to set s for 0 and solve for t
Is that 0 =-t^2 plus 5t plus 500
S is for distance
t is for time
(A) 0= -t^2 + 5(t) + 500 <<< This is a quadratic equation solve it by your favorite method

how do I find out where to put everything.I don't have a example to o by help!!
 

Subhotosh Khan

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jamekadiva said:
We have not been over the vertex I'M LOST!!
A generalized equation of parabola is:

\(\displaystyle y \, = \, (x-h)^2 \, + \, k\)

where (h,k) is the co-ordinate of the vertex.
 

mmm4444bot

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Have you learned about "completing the square"?

We can complete the square to go from

y = -t^2 + 5t + 500

to

y = (t - h)^2 + k

Are you familiar with the quadratic equation y = at^2 + bt + c?

The x-coordinate of the vertex is -b/(2a).

Find the corresponding y-coordinate by using one of the equations.

 

jamekadiva

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I'm in 14 and in the 9th grade. We have not been over that.
 

jamekadiva

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Thank You all for your help. I still can't git it but I'm still here. :D
 

Denis

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jamekadiva said:
I'm in 14 and in the 9th grade. We have not been over that.
Ok...but we have no idea WHAT you're able to do :shock:

Can you solve this equation for x :
x^2 - 2x - 35 = 0

If not, then there's LOTS of stuff you haven't "been over".
 
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