Part a: Option D 18 feet per second
Part b: increasing
Solution:
Height [tex]h(t)=-16 t^{2}+50 t+3[/tex]
Part a: To find the average rate of change for h(t) between t = 0 and t = 2.
Substitute t = 0 in h(t).
[tex]h(0)=-16 (0)^{2}+50 (0)+3[/tex]
h(0) = 3
Substitute t = 2 in h(t).
[tex]h(2)=-16 (2)^{2}+50 (2)+3[/tex]
h(2) = 39
Average rate of change formula:
[tex]$=\frac{h(b)-h(a)}{b-a}[/tex]
Here, a = 0 and b = 2.
[tex]$=\frac{h(2)-h(0)}{2-0}[/tex]
[tex]$=\frac{39-3}{2}[/tex]
[tex]$=\frac{36}{2}[/tex]
= 18
Average rate of change = 18 feet per second
Option D is the correct answer.
Part b:
This means height of the ball is increasing for 0 < x < 2.
