SOLUTION: 1. What is the maximum height that a basketball will have if thrown up into the air from a height of six feet, with an initial speed of 20 feet per second? (Must use quadratic fo
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Question 1005442: 1. What is the maximum height that a basketball will have if thrown up into the air from a height of six feet, with an initial speed of 20 feet per second? (Must use quadratic formula).
2. After how many seconds does the basketball hit the ground? (Must solve by completing the square)
Thank you anyone for helping me.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
quadratic formula for height at time t is
s(t) = -gt^2 + vot + ho, where g is force of gravity, vo is the objects initial velocity, and ho is initial height
since your problem is in feet we use g = 16
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1) s(t) = -16t^2 + 20t + 6
this is the equation of a parabola that curves downward
the coordinate of a point on the parabola is given by (t, s), that is time t is the x coordinate and height s is the y coordinate,
we know that the t coordinate of the vertex is given by
t = -b/(2a) = -20 / (2 *(-16)) = 20/32 = 5/8 = 0.625
s(0.625) = -16*(0.625)^2 + 20*(0.625) + 6 = 12.25
maximum height of basketball is 12.25 feet
here is a graph of the basketball's flight
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2) height s is 0 when the basketball hits the ground
-16t^2 +20t +6 = 0
divide both sides of = by -16
t^2 -(20/16)t -6/16 = 0
reduce fractions
t^2 -(5/4)t -(3/8) = 0
add 3/8 to both sides of =
t^2 -(5/4)t = 3/8
note (1/2) of (5/4) = (5/8)
t^2 -(5/4)t +(25/64) = (3/8) + (25/64)
note that 3/8 = 24/64
(t - 5/8)^2 = 49/64
take square root of both sides of =
t - 5/8 = 7/8
t = 5/8 + 7/8 = 12/8 = 1.5
after 1.5 seconds the basketball hits the ground
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compare our answers to 1 and 2 to the graph of the basketball's flight
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