A = 48 ft/ s
B = 960 ft
C Find the t coordinate of the vertex = - 48 / (2 * -16) = 48/32 = 3/2 sec
Plug this back into the function
-16(3/2)^2 + 48(3/2) + 960 = 996 ft = maax height
D ( from C ) = 3/2 sec
E The object will hit to ground when h = 0....so
-16t^2 + 48t + 960 = 0 divide through by -16
t^2 - 3t - 160 = 0 complete the square on t
t^2 -3t + 9/4 = 160 + 9/4
(t - 3/2)^2 = 649/4 take both roots and we have that
t = sqrt (694)/2 + 3/2 = 14.67 sec or t = -sqrt (694/)/2 +3/2 = -11.67 sec
Take the positive answer = 14.67 sec
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