The height, in metres, of an object that has fallen from a height of 180 m is given by the position function s(t)=-􏰀5t^2 􏰂+180, where t 􏰃>0 and t is in seconds.

a. Find the average velocity during each of the first two seconds.
b. Find the velocity of the object when t 􏰀=4.
c. At what velocity will the object hit the ground?

a. The average velocity during the first two seconds can be found by calculating the change in position divided by the change in time.

For the first second (t = 1), the position function is s(1) = -5(1)^2 + 180 = 175 m.
For the second second (t = 2), the position function is s(2) = -5(2)^2 + 180 = 160 m.

Therefore, the change in position is 175 - 180 = -5 m.

The change in time is 2 - 1 = 1 second.

Average velocity = (change in position) / (change in time) = -5 m / 1 s = -5 m/s.

For the first second, the average velocity is -5 m/s.

For the second second (t = 2), the change in position is 160 - 175 = -15 m.

The change in time is 2 - 1 = 1 second.

Average velocity = (change in position) / (change in time) = -15 m / 1 s = -15 m/s.

For the second second, the average velocity is -15 m/s.

b. To find the velocity of the object when t = 4, we need to find the derivative of the position function with respect to time.

The derivative of s(t) = -5t^2 + 180 is v(t) = -10t.

Plugging in t = 4, we get v(4) = -10(4) = -40 m/s.

Therefore, the velocity of the object when t = 4 is -40 m/s.

c. The object will hit the ground when the position is 0.

Setting s(t) = -5t^2 + 180 = 0, we can solve for t.

-5t^2 = -180
t^2 = 36
t = ±6

Since t must be positive, the object will hit the ground at t = 6 seconds.

Now, to find the velocity at this time, we can plug t = 6 into the derivative of the position function:

v(6) = -10(6) = -60 m/s.

Therefore, the object will hit the ground with a velocity of -60 m/s.

a. To find the average velocity during each of the first two seconds, we need to find the displacement of the object during each of those time intervals.

We can calculate the displacement by finding the value of s(t) at the end of each time interval and subtracting the value of s(t) at the beginning of the interval.

For the first second (t = 0 to t = 1):
s(1) = -5(1)^2 + 180 = -5 + 180 = 175 m
s(0) = -5(0)^2 + 180 = 0 + 180 = 180 m

Displacement = s(1) - s(0) = 175 - 180 = -5 m

Average velocity = Displacement / Time interval
Average velocity for the first second = -5 / 1 = -5 m/s

For the second second (t = 1 to t = 2):
s(2) = -5(2)^2 + 180 = -20 + 180 = 160 m
s(1) = -5(1)^2 + 180 = -5 + 180 = 175 m

Displacement = s(2) - s(1) = 160 - 175 = -15 m

Average velocity for the second second = -15 / 1 = -15 m/s

Therefore, the average velocity during the first second is -5 m/s and during the second second is -15 m/s.

b. To find the velocity of the object when t = 4, we need to find the derivative of the position function with respect to time, which will give us the velocity function.

v(t) = s'(t) = d/dt(-5t^2 + 180)

Using the power rule for derivatives, we get:

v(t) = -10t

Now, substitute t = 4 into the velocity function:

v(4) = -10(4) = -40 m/s

The velocity of the object when t = 4 is -40 m/s.

c. The object hits the ground when s(t) = 0.

Let's set s(t) = 0 and solve for t:

-5t^2 + 180 = 0
-5t^2 = -180
t^2 = 36
t = √36
t = 6

Therefore, the object hits the ground when t = 6 seconds.

To find the velocity at which the object hits the ground, we need to find the velocity function v(t) at t = 6:

v(t) = -10t

v(6) = -10(6) = -60 m/s

Therefore, the object hits the ground with a velocity of -60 m/s.

To find the average velocity during each of the first two seconds, we need to find the change in position (height) divided by the change in time.

a. Average velocity during the first second:
We know that t = 0 seconds corresponds to the object starting its fall, and t = 1 second corresponds to the end of the first second. We need to calculate the change in position from t = 0 to t = 1 and divide it by the change in time.

s(1) = -5(1)^2 + 180 = -5 + 180 = 175 meters
s(0) = -5(0)^2 + 180 = 180 meters

Change in position = s(1) - s(0) = 175 - 180 = -5 meters
Change in time = 1 - 0 = 1 second

Average velocity during the first second = Change in position / Change in time = -5 meters / 1 second = -5 m/s

b. Average velocity during the second second:
Now, we need to find the change in position from t = 1 to t = 2 and divide it by the change in time.

s(2) = -5(2)^2 + 180 = -20 + 180 = 160 meters
s(1) = -5(1)^2 + 180 = -5 + 180 = 175 meters

Change in position = s(2) - s(1) = 160 - 175 = -15 meters
Change in time = 2 - 1 = 1 second

Average velocity during the second second = Change in position / Change in time = -15 meters / 1 second = -15 m/s

b. To find the velocity of the object when t = 4, we need to take the derivative of the position function with respect to time.

s(t) = -5t^2 +180
v(t) = d(s(t))/dt = -10t

Substituting t = 4 into the velocity function:
v(4) = -10(4) = -40 m/s

c. The object will hit the ground when its height is zero, so we need to find the time at which s(t) = 0.

-5t^2 + 180 = 0
-5t^2 = -180
t^2 = 36
t = ±6

The object will hit the ground at t = 6 seconds.

To find the velocity at which the object hits the ground, we can substitute t = 6 into the velocity function.

v(6) = -10(6) = -60 m/s

Therefore, the object will hit the ground with a velocity of -60 m/s.

s(t) = 180-5t^2

(a) avg velocity is total distance/total time.
s(2) = 180-5*4 = 160
160m/2s = 80 m/s

v(t) = ds/dt = -10t
v(4) = -40

180-5t^2 = 0
t = 6
That is, at t=6, the height is zero -- it hit the ground.
Now you just plug 6 into v(t)