Thank you for posting your question here. I hope the answer below will help.
Vo=110 feet per second
<span>ho=2 feet </span>
<span>So, h(t) = -16t^2 +110t +2 </span>
<span>Take the derivative: h'(t) = 110 -32t </span>
<span>The maximum height will be at the inflection when the derivative crosses the x-axis aka when h'(t)=0. </span>
<span>So, set h'(t)=0 and solve for t: </span>
<span>0 = 110 -32t </span>
<span>-110 = -32t </span>
<span>t=3.4375 </span>
<span>t=3.44 seconds </span>
<span>90.0550458716 is the answer</span>
1.
Height h = 240 feet
We insert it into formula.

Now we solve this for t:

We have two answers as the rocket will pass this height on it's way up and down.
2.
Height h = 0 feet
We insert it into formula.

Now we solve this for t:

We do not consider solution -1 as the time can not be negative. Our solution is 5s.
3.
Height at bottom of hill h = 0 feet
We insert it into formula.
Answer:
$53.07
Step-by-step explanation:
49.95x6.25%=3.12
49.95+3.12=53.07