A batter hits a baseball straight up into the air. The height of the ball in metres above the ground is given by h(t)=-4.9t^2+23t+2, where t is the number of seconds after the ball is hit.

a) What is the average rate of change in the height of the ball over the interval 1.5<=t<=2?
b) What is the instantaneous rate of change in the height of the ball at 2 seconds?

a)h(2)-h(1.5)/2-1.5

h(2)= -4.9(2)^2+23(2)+2
= 28.5

h(1.5)= -4.9(1.5)^2+23(1.5)+2
= 25.475

28.5-25.475= 2.925
2-1.5= 0.5

2.925/0.5= 5.85

Therefore the average rate of change is 5.85m/s

a) Well, the average rate of change is like the average dance move at a clown party. It's all about the difference between the final height and the initial height, divided by the difference in time. So, let's calculate it!

At t = 1.5 seconds, the height of the ball is h(1.5) = -4.9(1.5)^2 + 23(1.5) + 2.

At t = 2 seconds, the height of the ball is h(2) = -4.9(2)^2 + 23(2) + 2.

Now, we just need to subtract them and divide by the difference in time (2 - 1.5). Math is a great way to channel your inner clown, isn't it?

b) Oh, the instantaneous rate of change! It's like catching a glimpse of a clown's funny face when they're mid-juggle. To find that, we need to calculate the derivative of h(t) and then evaluate it at t = 2 seconds.

The derivative of h(t) = -4.9t^2 + 23t + 2 is h'(t) = -9.8t + 23.

Now, let's substitute t = 2 into h'(t): h'(2) = -9.8(2) + 23.

So, the instantaneous rate of change at 2 seconds is h'(2). Just like a clown juggling flaming torches, it's all about being balanced and calculated!

To find the average rate of change, we need to calculate the difference in height at the beginning and end of the interval, and divide it by the difference in time.

a) Average rate of change:

To find the height at t = 1.5 seconds, substitute t = 1.5 into the equation:
h(1.5) = -4.9(1.5)^2 + 23(1.5) + 2

To find the height at t = 2 seconds, substitute t = 2 into the equation:
h(2) = -4.9(2)^2 + 23(2) + 2

Now we can find the average rate of change:
Average rate of change = (h(2) - h(1.5)) / (2 - 1.5)

b) Instantaneous rate of change:

To find the instantaneous rate of change at 2 seconds, we need to find the derivative of the function h(t) with respect to t, and then evaluate it at t = 2.

To find the derivative, differentiate each term of the equation:
h'(t) = -4.9(2t) + 23

Now substitute t = 2 into the derivative:
h'(2) = -4.9(2)(2) + 23

Therefore:
a) The average rate of change in the height of the ball over the interval 1.5 <= t <= 2 is (h(2) - h(1.5)) / (2 - 1.5).
b) The instantaneous rate of change in the height of the ball at 2 seconds is h'(2) = -4.9(2)(2) + 23.

To find the average rate of change in the height of the ball over the interval 1.5<=t<=2, we need to calculate the change in height and the change in time over that interval.

a) Average rate of change in height = (change in height) / (change in time)

1. Calculate the height of the ball at t = 1.5 seconds:
h(1.5) = -4.9*(1.5)^2 + 23*(1.5) + 2

2. Calculate the height of the ball at t = 2 seconds:
h(2) = -4.9*(2)^2 + 23*(2) + 2

3. Calculate the change in height over the interval:
(change in height) = h(2) - h(1.5)

4. Calculate the change in time over the interval:
(change in time) = 2 - 1.5

5. Calculate the average rate of change in height:
Average rate of change = (change in height) / (change in time)

b) To find the instantaneous rate of change in the height of the ball at 2 seconds, we need to find the derivative of the height function h(t) with respect to t.

1. Differentiate the function h(t) = -4.9t^2 + 23t + 2 with respect to t to find the instantaneous rate of change:
h'(t) = -9.8t + 23

2. Calculate the instantaneous rate of change at t = 2 seconds:
Instantaneous rate of change = h'(2) = -9.8(2) + 23

So,

a) The average rate of change in the height of the ball over the interval 1.5<=t<=2 is (change in height) / (change in time).
b) The instantaneous rate of change in the height of the ball at 2 seconds is h'(2) = -9.8(2) + 23.

just for fun, let's see when v = 0

9.8 t = 23
t = 2.3 seconds at top so our problem is all on the way up.
so
h(2) = -4.9(4) + 23(2) + 2
h(1.5)=-4.9(2.25)+23(1.5) + 2

h(2)-h(1.5) = -4.9(4-2.25) + 23(.5)
= 4.5
so
v average = 4.5/.5 = 9 m/s

part b
well I already differentiated to find velocity
v = dh/dt = -9.8 t + 23
at t = 2
v = 3.4 m/s getting close to top :)