217k views
5 votes
A rocket carrying fireworks is launched from a hill 60 feet above a lake. The rocket will fall into the lake after exploding at its maximum height. The rocket's height above the surface of the lake is given by the function h(t) = -16t2 + 96t + 60. Part A: What is the maximum height reached by the rocket? Part B: What is the height of the rocket after 2 seconds?

1 Answer

2 votes

Answer:

A) 204m

B) 188m

Explanation:

Given the rocket's height above the surface of the lake given by the function h(t) = -16t^2 + 96t + 60

The velocity of the rocket at its maximum height is zero

v = dh/dt = -32t + 96t

At the maximum height, v = 0

0 = -32t + 96t

32t = 96

t = 96/32

t = 3secs

Substitute t = 3 into the modeled function to get the maximum height

h(3) = -16(3)^2 + 96(3) + 60

h(3) = -16(9)+ 288 + 60

h(3) = -144+ 288 + 60

h(3) = 144 + 60

h(3) = 204

Hence the maximum height reached by the rocket is 204m

Get the height after 2 secs

h(t) = -16t^2 + 96t + 60

when t = 2

h(2) = -16(2)^2 + 96(2) + 60

h(2) = -64+ 192+ 60

h(2) = -4 + 192

h(2) = 188m

Hence the height of the rocket after 2 secs is 188m

User Alexander Wallin
by
8.3k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories