Question 1036786
The height of a projectile t seconds after being launched with an initial
velocity of 96 ft/sec from an initial height of 200 ft above the ground is given by the
following formula:
s(t) = -16t^2 + 96t + 200

(a) At what moment in time is the height of the projectile the greatest?
"moment in time" is redundant.
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(a) At what time is the height of the projectile the greatest?
It's the vertex of the parabola, at t = -b/2a
t = -96/-32 = 3 seconds.
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(b) At what time does the projectile return to the ground?
When s(t) = 0.
-16t^2 + 96t + 200 = 0
-2t^2 + 12t + 25 = 0
*[invoke solve_quadratic_equation -2,12,25]
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Ignore the negative solution.
t =~ 7.637 seconds
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(c) The slope of the line through points (t0, s(t0)) and (t1, s(t1)) is the average velocity
on the time interval [t0, t1]. What is the average velocity of the particle from the
moment it was launched to the moment when it reaches its greatest height?
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t0 = 0, t1 = 3
s0 = 200, s1 = s(3) = -16*9 + 96*3 + 200 = 344
--> (344-200)/3 = 48 ft/sec