SOLUTION: A programmer is writing the code for a new interactive basketball game. As a result, she is using quadratic relations to model the path of the ball. During the game, when a ball is

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Question 683994: A programmer is writing the code for a new interactive basketball game. As a result, she is using quadratic relations to model the path of the ball. During the game, when a ball is shot, the path it follows is modeled by the quadratic relation,h= -0.2d^2 + 3d + 6 , where h represented the height of the ball above the ground and d represented the distance of the ball from the shooter. Both distances are measured in feet.
How high was the ball when the shooter shot it?
What was the maximum height obtained by the ball?
A rim of a basketball net is 10 feet high. For what horizontal distance was the ball above the rim of the basketball net?
Please help for my review coming up ):
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
How high was the ball when the shooter shot it?

h=+-0.2d%5E2+%2B+3d+%2B+6

h=+-0.2%280%29%5E2+%2B+3%280%29+%2B+6 ... Plug in d = 0 (since the ball is zero ft away from the player when it is shot)

h=+-0.2%280%29+%2B+3%280%29+%2B+6+

h=+0+%2B+0+%2B+6+

h+=+6

So the ball is 6 ft off the ground when the shooter shot it.

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What was the maximum height obtained by the ball?

To answer this question, we need to find the vertex

First replace all 'h' terms with y. Then replace all 'd' terms with 'x'

Now find the x coordinate of the vertex

x+=+-b%2F%282a%29

x+=+-3%2F%282%28-0.2%29%29

x+=+7.5

Then plug this in to find the y coordinate of the vertex

y+=+-0.2d%5E2+%2B+3d+%2B+6

y+=+-0.2%287.5%29%5E2+%2B+3%287.5%29+%2B+6

y+=+-0.2%2856.25%29+%2B+3%287.5%29+%2B+6

y+=+-11.25+%2B+22.5+%2B+6

y+=+17.25

The y coordinate of the vertex represents the highest point.

So the max height is 17.25 ft.

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A rim of a basketball net is 10 feet high. For what horizontal distance was the ball above the rim of the basketball net?


Plug in h = 10 and solve for d

h+=+-0.2d%5E2+%2B+3d+%2B+6

10+=+-0.2d%5E2+%2B+3d+%2B+6

0+=+-0.2d%5E2+%2B+3d+%2B+6+-+10

0+=+-0.2d%5E2+%2B+3d+-+4

-0.2d%5E2+%2B+3d+-+4+=+0

10%28-0.2d%5E2+%2B+3d+-+4%29+=+10%2A0

-2d%5E2+%2B+30d+-+40+=+0

-1%28-2d%5E2+%2B+30d+-+40%29+=+-1%2A0

2d%5E2+-+30d+%2B+40+=+0

2%28d%5E2+-+15d+%2B+20%29+=+0

d%5E2+-+15d+%2B+20+=+0%2F2

d%5E2+-+15d+%2B+20+=+0

Now use the quadratic formula to solve for d

d+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

d+=+%28-%28-15%29%2B-sqrt%28%28-15%29%5E2-4%281%29%2820%29%29%29%2F%282%281%29%29 Plug in a+=+1, b+=+-15, c+=+20

d+=+%2815%2B-sqrt%28225-%2880%29%29%29%2F%282%29

d+=+%2815%2B-sqrt%28145%29%29%2F2

d+=+%2815%2Bsqrt%28145%29%29%2F2 or d+=+%2815-sqrt%28145%29%29%2F2

d+=+13.520797 or d+=+1.479203

So the distance is either d+=+13.520797 or d+=+1.479203, which means that the distance it travels about 13.520797 - 1.479203 = 12.041594 feet horizontally when the ball is above the 10 ft rim.