Aball is thrown straight up with a initial velocity of 56 feet per second . the height , h , of the ball t seconds after it is thrown is given by the formula h(t)=56t-16t^2 . what is the maximum height of the ball?

Formula: H(t) = 56t – 16t^2  

H(t) = - 16t^2 + 56t

A.      What is the height of the ball after 1 second? H (1) = 56(1) – 16(1) ^2 = 40 pt.

B.      What is the maximum height? X = - (56)/2(- 16) = 1.75 sec h (1.75) = 56(1.75) – 16(1.75) ^2 h (1.75) = 49ft.

C.      After how many seconds will it return to the ground? – 16t^2 + 56t = 0   - 8t =0 t = 0

-          8t (2 + - 7) = 0     2t – 7 = 0  t = 7/2   Ans: 3.5 seconds

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.