A ball is thrown vertically upward with an initial velocity of 80 ft. per second. The distance s (in ft.) of the ball from the ground after t seconds is given by the function s(t)=80t-16t^2. [spoiler]a) in what time interval is the ball more than 96ft above the ground? s(t) = 96 ft 96 = 80t - 16t^2 96 = -16 (t^2-5t) -6= (t^2 - 5t) t^2 - 5t + 6 (t-3)(t-2) t= 3 and t= 2 b) What is the maximum height of the ball ? ** Hindi ako sure s(t) = 80t - 16t^2 or --> s(t)= -16t^2 + 80t standard formula for vertex: f(x) = a(x-h)^2 +k simplified: f(x)= a (x^2 - 2xh + h^2) finding the constant (aka h^2): -2xh/2x = -h then i-square. XD -> h^2 vertex is (h,k) s(t) = -16 (t^2 - 5t) <-- maglalagay ng constant bale... -5t/2t = (-5/2)^2 = 25/4 equation: s(t) = -16(t^2 - 5t +25/4) pero dpat even equation, maglalagay din ng 25/4 sa kbila. kasama din yung (-16) s(t) - (16)(25/4) = -16(t^2 - 5t +25/4) s(t) -100 = -16(t^2 - 5t +25/4) *transpose. s(t) = -16(t^2 - 5t +25/4) + 100 * factor out expression s(t) = -16 (t-5/2)^2 + 100 --> standard equation --> f(x) = a(x-h)^2 +k vertex; (h,k)--> (5/2, 100) max height; 100ft[/spoiler] lift lang. imyall. |: