Question 211169
This is a summer math problem I was assigned. At t=0 seconds Vikki stood on the roof of a buildng and threw a ball in the air. The height of the ball at t seconds is gven by the formula H(t)=-16t2*+96t+56. 
A:Answered
B:When is the ball at its highest point?
There's more than one way to find that.  One is to find the time when the ball returns to 56 feet, and the max height it 1/2 of that time, ie, it takes the ball the same amount of time going up as it does coming down.
H(t)=-16t2*+96t+56
56 = -16t2*+96t+56
-16t^2 + 96t = 0
t*(-16t + 96) = 0
t = 0 (the launch)
t = 6 seconds (on its way back down)
So the time at apogee (max height) is 3 seconds.
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C:To what height does the ball travel?
h(3) = -16*9 + 96*3 + 56
h(3) = 200 feet (apogee at t = 3 seconds)
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D:When does the ball strike the ground?
h(t) = 0 is ground level.
-16t2*+96t+56 = 0
-2t^2 + 12t + 7 = 0
*[invoke solve_quadratic_equation -2,12,7]
time of impact is 6.5355... seconds.  Ignore the negative number.
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