calc~~~ help pls!! ^^?

Question

given the position function h(t)=-16t^2 + 85t +470 answer the following questions.

a) explain the meaning of the numbers and units of this position function.

b) find the height of the object after 3 seconds.

c) find the average velocity of the object over the first 3 seconds.

d) find h (3) and explain the meaning of this computation.

e) find the veloctiy of the object the moment it hits the ground

Answer 1

a) t is time elapsed, h(t) is height of the object in parabolic form when the object is thrown out.
Since h(t) = -16t^2 +85t + 470, the object start at level 470 m from ground floor.
Assume t is second, h is meter

b) After 3 seconds,
height of the object = -16(3)^2 + 85(3) +470 = 581 m.

c) After 3 seconds,
h(t) = -16t^2 +85t + 470
dh/dt = -32t + 85
average velocity of object, dh/dt = -32(3) +85 = -11 m/s
(negative means object is go down)

d) h(3) = -16(3)^2 + 85(3) +470 = 581 m.
h(3) is the position of the object after 3 seconds.

e) When object hit ground,
h(t) = -16t^2 +85t + 470 = 0
16t^2 -85t -470 = 0
(t-8.7) (t+3.38) = 0
t cannot negative
t = 8.7 s

Velocity of object hitting ground
dh(8.7)/dt = -32(8.7) + 85 = -193.4 m/s

Answer 2

a) The 16 represents 1/2 of the acceleration due to gravity, 32 ft/sec. The 85 is the initial velocity, and the 470 is the initial position (height above the ground).

b) After 3 seconds the object will be:

h(3) = -16(3)² + 85(3) + 470 = 581 feet above the ground.

c) The velocity is the derivative of the position:

v(t) = -32t + 85

v(3) = -32(3) + 85 = -11

The average velocity is (85 + (-11))/2 = 37 ft/s

d) Not sure what the difference is between this and part b?

e) It hits the ground when h(t) = 0.

0 = -16t² + 85t + 470

t = (-85 ± √(85² - 4(-16)(470))/(2(-16))
t = (-85 - 193.145)/-32
t = 8.692 s

v(8.692) = -32(8.692) + 85 = -192.145 ft/s

if I didn't mess up.

Answer 3

(a) The question is ambiguous. "470" carries the unit of meters (or some other distance) and represents the initial position. "85" carries the unit of meters per second and represents the initial velocity. "-16" carries the unit of meters per second squared and is half of the (constant) acceleration.

(b) h(3) = -16*3^2 + 85*3 + 470 = -144 + 255 + 470 = 581 meters

(c) h'(t) = 85 - 32t. Since this is linear in t, we can compute its average value over an interval by just averaging the values at the two endpoints of the interval. h'(0) = 85; h'(3) = -11 ==> average value = 37 meters per second

(d) redundant to (b)

(e) Solve h(t) = 0. The positive solution is t=8.692. Plug this into h'(t): h'(8.692) = 85 - 32(8.692) = -193.144 meters per second.