A model rocket is launched upward at a speed of 96 feet per second from a platform 7 feet above the ground. The path of the rocket can be modeled by the projectile equation, h(t)=-16t^2+v₀t+h₀
1. What is the maximum height the rocket will reach?
2. What is the average rate of change from the initial height to the maximum height?

151 ft

48 ft/s

Step-by-step explanation:

h(t)=-16t^2+v₀t+h₀

v₀ = 96  and

h₀ = 7

h(t)=-16t^2+96t+7

The maximum height is at the vertex

The x values of the vertex is at

x = -b/2a = -96/ (2*-16) = -96/-32 =3

Substitute this into the function to find the maximum height

h(3) = -16 ( 3)^2 +96*3+7

    -16*9 +288+7

     151

The maximum height is 151 ft

Average rate of change from 0 to 3

h(3) - h(0)

3-0

151 -7

3

144/3

48ft/s

step-by-step explanation: