Question

a. Considering the context of the problem, what is an appropriate domain for h(t)?

ht? Explain your reasoning.

 

 ≤t≤

≤t≤ ; an appropriate domain is all times from the launch to when the watermelon hits the ground or  seconds to  seconds.

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Part B

Fill in the blank question.b. Use the graph of h(t)

ht to find the maximum height of the watermelon. When does the watermelon reach the maximum height? Explain your reasoning.

 

 feet at  seconds; the graph of h(t)

ht has a maximum at (, ), which represents the time at which the watermelon reaches it maximum height.

Answer 1

Height of the watermelon above the ground is modeled by the quadratic function

$$h(t)=-16t^2+v_0t+h_0,$$

for $t$ seconds and $h$ feet.

$$t\geq0, h\geq0$$

$$-16t^2+v_0t+h_0=0$$

$$D=(v_0)^2-4(-16)(1)=v_0^2+64>0$$

$$t=\dfrac{-v_0\pm\sqrt{v_0^2+64}}{2(-16)}=\dfrac{v_0\pm\sqrt{v_0^2+64}}{32}$$

Since $t\geq0,$ we take

$$t=\dfrac{v_0+\sqrt{v_0^2+64}}{32}$$

a. Domain for $h(t)$

$$0\leq t\leq\dfrac{v_0+\sqrt{v_0^2+64}}{32}$$

b.

$$h(t)=-16t^2+v_0t+h_0$$

$$h'(t)=(-16t^2+v_0t+h_0)'=-32t+v_0$$

Find the critical number(s)

$$h'(t)=0=>-32t+v_0=0$$

$$t=\dfrac{v_0}{32}, v_0>0$$

$$h(\dfrac{v_0}{32})=-16(\dfrac{v_0}{32})^2+v_0(\dfrac{v_0}{32})+h_0=\dfrac{v_0^2}{64}+h_0$$

The watermelon reaches the maximum height with value of $\dfrac{v_0^2}{64}+h_0$ feet at time $t=\dfrac{v_0}{32}$ seconds.