The height h, in feet, of a ball that is released 4 feet above the ground with an initial velocity of 80 feet per second is a function of the time t, in seconds, the ball is in the air and is given by

h(t)=-16t^2+80t+4, 0 < t < 5.04

a. find the height of the ball above the ground 2 seconds after it is released.

b. find the height of the ball above the ground 4 seconds after it is released.

show work!

Part a)

Height of ball after 2 seconds it is released is

h(2)=-16\times 2^2+80\times 2+4=100feet

Part b)

Height of ball after 4 seconds it is released is

h(4)=-16\times 4^2+80\times 4+4=68feet

Step-by-step explanation:

7

step-by-step explanation:

(0,,,0)

step-by-step explanation:

the table of x and y values is shown in the attached pic(part of question)

we have to decide according that graph

given points are (0, 0) (–1, 0), (2, 0)

when we graph these points on graph they show an increasing line as shown in the attached pic