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?

Step-by-step 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

22 dimes and 9 nickels

step-by-step explanation:

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