Interpretation of the Derivative — Question 8

Question 8 - Solution

We are given: $$h(t) = -16t^2 + 64t + 80$$

(a) Velocity Function:

The velocity is the first derivative of height: $$v(t) = h'(t) = -32t + 64$$

$$\boxed{v(t) = -32t + 64}$$

(b) Time of Maximum Height:

The ball reaches its maximum height when the velocity is zero: $$-32t + 64 = 0 \Rightarrow t = 2$$

$$\boxed{t = 2 \text{ seconds}}$$

(c) Maximum Height:

Plug $t = 2$ into the original height function: $$h(2) = -16(2)^2 + 64(2) + 80 = -64 + 128 + 80 = 144$$

$$\boxed{144 \text{ feet}}$$

(d) Acceleration Function:

Acceleration is the derivative of velocity: $$a(t) = v'(t) = -32$$

$$\boxed{a(t) = -32 \text{ ft/s}^2 \text{ (constant acceleration due to gravity)}}$$