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.

Question

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?

Answer 1

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 :)

Answer 2

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