Question 646624
For a typical basketball shot, the balls height (in feet) will be a function of time in flight (in seconds), models by any equations such as h= -16t^2 + 40t 
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What is the maximum height of the ball?
h= -16t^2 + 40t
This is an equation of a parabola that opens downward. Curve has a maximum.
Its standard form: y=(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex. k=maximum
complete the square:
h= -16(t^2-40t/16
h= -16^2 +5t/2
h=-16(t^2-5t/2+25/16)+25
h=-16(t-5/4)^2+25
maximum height=25 ft
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When will the shot reach the height of the basket (10 feet)?
10= -16t^2 + 40t
16t^2-40t+10=0
solve for t by following quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
a=16, b=-40, c=10
ans: 
on the way up≈0.28 sec
on the way down≈2.22 sec
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When will the ball hit the floor, if it is missed the basket entirely?
0= -16t^2 + 40t
16t^2-40t=0
t(16t-40)=0
t=0 (reject, t>0)
or
16t-40=0
t=40/16=5/2=2.5 sec