Question 1029642: A bouncy ball is thrown vertically upward from the ground, after 2 seconds it attains max height of 120 feet, and after 4 seconds it hits the ground.
On a graph I have my vertex as (2, 120).
I am trying to find the quadratic equation to represent the path of the ball. t is time of ball in air and s is height of ball at time t.
I have figured out the equation to be -16t^2+96t=s, then I plug in to the equation t= (-b ± √(b^2-4 (a)(c) ))/2a
My answers are sorta making sense in relation to my graph, but we are asked to give height of ball at 4 seconds & 6 seconds. If ball was already on ground at 4 seconds, why are they asking at 6 too? Not positive I am on the right track.
Thank you!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A bouncy ball is thrown vertically upward from the ground, after 2 seconds it attains max height of 120 feet, and after 4 seconds it hits the ground.
On a graph I have my vertex as (2, 120).
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y = ax^2 + bx + c
Since the vertex is at (2,120), -b/(2a) = 2 ; = b = -4a
120 = a(4) -4a(2) + c
120 = -4a + c
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Since (4,0) is a point, a(16) + b(4) + c = 0
16a + 4b + c = 0
16a + 4(-4a) + c = 0
c = 0
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So, 120 = -4a
a = -30
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Therefore:: y = -30x^2 + 120x
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We are asked to give height of ball at 4 seconds & 6 seconds.
At 4 seconds:
f(4) = -30(16) + 120(4) = -480 + 480 = 0 (at 4 sec it is on the ground).
But you knew that.
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At 6 seconds:
Well, it is a bouncy ball but you have no info as to the % of rebound,
so ??????
Cheers,
Stan H.
If ball was already on ground at 4 seconds, why are they asking at 6 too? Not positive I am on the right track.
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