Help quick!!!!!!

Question

A worker accidentally drops a hammer from the scaffolding of a tall building. The worker is 300 feet above the ground. As you answer the following, recall that an object falls 16t^2 feet in t seconds (assuming negligible air resistance).

(a) How far above the ground is the hammer after falling for one second? for two seconds? Write a formula that expresses the height h of the hammer after it has fallen for t seconds.

(b) How many seconds does it take the hammer to reach the ground? How many seconds does it take for the hammer to fall until it is 100 feet above the ground?

(c) By plotting some data points and connecting the dots, sketch a graph of h versus t. Is your graph is a picture of the path followed by the falling hammer?

Answer 1

(a)

h(t) = -16t² + 300

h(1) = -16(1)² + 300

h(2) = -16(2)² + 300

(b)

0 = -16t² + 300

100 = -16t² + 300

(c)

You draw the graph.

Answer 2

h(t) =-16t^2 +300

h(1) = 300-16 = 284 feet after 1 second

h(2) = 300-32 = 268 feet after 2 seconds

it hits the ground when h(t)=0 = -16t^2 +300

t^2 = 300/16 = 18.75

t = sqr18.75 = about 4.33 seconds to reach ground zero

h(t) = 100 = -16t^2+300 where t= time when it reaches 100 feet

t^2 = 200/16= 12,5

t = sqr12.5

t= about 3.54 seconds

graphically it's an downward opening parabola with vertex = (0,300)= maximum point = y intercept

x intercept = about 4.33 or the point (4.33, 0)

points on the graph are (2,268), (1,284), (4.33,), (0,300) and (3.54, 100) connect them with a smooth parabolic curve. Plot them in quadrant I, with time on the horizontal axis and height on the vertical axis