Please help i'm not sure how to do this

Question

Julia launched a rocket vertically upward from the top of a cliff. The height of the rocket above the base of the cliff is approximated by the model h=-5t^2+30t+20, where "h" is the height in meters, and "t" is the time in seconds

A) how high is the cliff?

B) After how many seconds does the rocket hit the ground?

C) When does the maximum height occur?

D) What is the maximum height?

E) When is the rocket at 45m from the ground

Answer 1

Hey Luke!

a) Since the rocket starts on the top of the cliff, plug in 0 seconds for t into the equation. This gives you that the cliff is 20 meters high. (-5(0)²+30(0)+20 = 20)

b) Since this equation is not factorable to find the roots (or zeros), we can use the quadratic formula to find that the height equals 0 at (-30-√(30²-(4*-5*20)))/(2*-5) seconds or approximately 6.606 seconds after launch. (The negative answer from the quadratic formula cannot be applied in this scenario.)

c) The maximum height can be represented by the vertex. A shortcut to finding the x-value of the vertex is -b/2a. This means that the maximum height is reached at -30/-10 = 3 seconds after launch.

d) Using the information from part c, we can plug in 3 seconds for t into our equation. This gives us that the maximum height is 65 meters. (-5(3)²+30(3)+20 = 65)

e) Since we are looking for the time it takes to reach a certain height, we can plug in 45 meters for h in our equation. This gives us -5t²+30t+20 = 45. We can rearrange this equation to get -5t²+30t-25 = 0. From here we can factor the equation to get our zeros. The equation can be factored to become -5(t²-6t+5) = -5(t-5)(t-1). This means that our zeros occur at 1 and 5 seconds. Therefore, our rocket reaches 45 meters at 1 second after launch and 5 seconds after launch.

Answer 2

h = -5t²+ 30t + 20

A) Height of cliff: At t = 0 (before launch) h = 0 + 0 + 20 = 20 meters

B) After how many seconds does rocket hit the ground? That happens when h = 0, since h is defined as height above the base of the cliff.

-5t² +30t + 20 = 0 Not easily factored, so use quadratic equation. There are websites (Symbolab) that calculate for you as well. Answer is 3 ± √13 sec. Use the positive value. Also if you graph it (DESMOS Graphing Calculator), you'll see where the graph crosses the x axis (ht = 0)

C) When does max. ht. occur? That happens at the vertex of the parabola. You can set the derivative to 0 and solve for t. Or, use -b/2a to find x value for the vertex. -30/[2(-5)] = -30/-10 = 3 sec. You can see that on the graph as well.

D) What is max. ht. Set t = 3 and solve for h.

E) When is rocket 45 meters from the ground. h = 45 so -5t² + 30t + 20 = 45

Need to factor -5t² +30t - 25 = 0. Can use "ac" method. There are 2 answers (one while going up, the other when coming back down). You can check your answer by looking at the graph.