Question

The height (in feet) of objects launched from a

pirate ship's rocket launcher can be modeled by

the equation h = -16t? + 180t + 18 where t is the

time in seconds.

*Find the maximum height a launched item

could reach. (Round to two decimal places).

*After how many seconds will an item hit the

water after being launched? (Round to nearest

whole number).

Please someone help me

Answer 1

i) 524.25feet

ii) 11secs

Explanation:

1) Given the height of the rocket modelled by the equation

h = -16t² + 180t + 18 where

t is in seconds

At the maximum height, the velocity of the object is zero.

v(t) = dh/dt = 0

dt/dt = -32t + 180

0 = -32t + 180

32t = 180

t = 180/32

t = 5.625secs

Get the maximum height

h = -16t² + 180t + 18

h = -16(5.625)² + 180(5.625) + 18

h = -16(31.640625)+ 180(5.625) + 18

h = -506.25+1,012.5+18

h = 524.25feet

Hence the maximum height a launched item could reach is 524.25feet

ii) The item will hit the water at h = 0

Substitute inti the expression

h = -16t² + 180t + 18

0 = -16t² + 180t + 18

16t² - 180t - 18 = 0

Divide through by 2

8t² - 90t - 9 = 0

t = 90±√90²-4(8)(-9)/2(8)

t = 90±√8100+288/16

t = 90±√8,388/16

t = 90±91.586/16

t = 90+91.586/16

t = 181.586/16

t = 11.35sec

Hence the item hit the water after 11secs (to nearest whole number)