Question 1005442
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
{{{graph(300,200,-2, 5, -2, 15, -16x^2+20x+6)}}}
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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