algebra 2 help

[Gabbyprincess2012]

Find the height, in feet, of the ball after 3 seconds in the air?

help?a ball is launched from a sling shot. its height, h(x), can be represented by a quadratic function in terms of time,x,in seconds.

After 1second, the ball is 121 feet in the air, after 2 seconds, it is 224 feet in the air.

 

[mmm4444bot]

help?

What kind of help would you like? In general, we do not provide worked-out solutions to exercises on these boards. We provide guidance, based on the information supplied.

Have you done or thought about anything, yet? Please share some information about your situation.

(The post titled "Read Before Posting" provides guidelines on how to ask for help here.)

Also, we would like to know whether you provided all of the given information. Specifically, did they tell you to assume anything? Is there a diagram?

Have you used a sling shot? How many feet above the ground do you suppose the ball is when x = 0 (i.e., the instant the sling shot releases the ball)?

 

[mmm4444bot]

Something seems goofy to me with the posted information because I find that the sling-shot operator must be located somewhere below the surface of the earth.

I'm assuming that the ball is shot vertically into the air (i.e., perpendicular to the ground, versus some other angle).

There is a common formula from physics to model the ball's height. When measured in feet, the height h(x) is given by:

h(x) = -16x^2 + Bx + C

where:

x is the number of elapsed seconds since release

h(x) is the corresponding height above the ground at time x

B is the ball's velocity (in feet per second) at release

C is the initial height (that is, the height of the ball at release)

You should plug-in the given pairs of values -- for x and for h(x) -- using this formula (substitute one pair at a time). Then simplify each result, to obtain two new equations. These two new equations form what we call a "system of two equations with two unknowns". The unknowns in your system are B and C.

Do you have any experience solving a system of two equations? Please let us know.

Solving the system gives you the values of B and C.

Once you have those, you have a formula for finding the ball's height h(x) at any number of elapsed seconds that make sense. The answer to the exercise is the value that you get for h(3).

Cheers :cool:

 

[HallsofIvy]

You say that the height is quadratic so it must be of the form \(\displaystyle h(t)= at^2+ bt+ c\). We will be able to find h at any height, and, in particular, at t= 3, if we can first find a, b, and c. To do that, we need three equations. We are told that "after 1 second, the ball is 121 feet in the air" so that h(1)= a(1)^2+ b(1)+ c= a+ b+ c= 121 and "after 2 seconds, it is 224 feet in the air" so that h(2)= a(2)^2+ b(2)+ c= 4a+ 2b+ c= 224. We need a third equation which would probably be the height at t= 0. Are you given information about that? Are we to assume that h(0)= 0 or was the slingshot initially some specifice height above the ground? Taking t= 0 we have h(0)= a(0)^2+ b(0)+ c= c.

If the initial height is, in fact, 0, we have the three equations c= 0, a+ b+ c= 121, and 4a+ 2b+ c= 224. With c= 0, the last two equations become a+ b= 121 and 4a+ 2b= 224. We can eliminate b from that by subtracting twice the first equation from the second: (4a+ 2b)- (2a+ 2b)= 224- 242 or 2a= -12 or a= -6. Then a+ b= 121 becomes -6+ b= 121 so that b= 127. You now have h(t)= -6t^2+ 121t. Now, what is h(3)?

If the initial height is something other than 0, just add that initial height to "h(3)".