Question 762872
Margie threw a ball upwards while standing on a platform 75 ft off of the ground. The trajectory after t seconds follows the equation: 
h(t) = –0.6t2 + 108t + 75.
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b.	How long will it take the ball reach its maximum height?
The equation is of a parabola.
The max ht is the vertex.
The vertex is on the axis of symmetry, t = -b/2a
t = -108/(2*-0.6) = 90 seconds
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a.	What will be the maximum height of the ball?
Max ht = h(90)
= -0.6*90^2 + 108*90 + 75
= 4935 ft
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c.	At what time will the ball hit the ground?
When h(t) = 0
h(t) = –0.6t^2 + 108t + 75 = 0
6t^2 - 1080t - 750 = 0
t^2 - 180t - 125 = 0
*[invoke solve_quadratic_equation 1,-180,-125]
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t =~ 180.69 seconds
Ignore the negative solution