Question

Can someone please solve these?

Answer 1

Explanation:

h = -6
`t^(2)` +20t + 4

I will use calculus, maybe that's not how you're supposed to do this

-12t +20 =0

12t = 20

t = 20 /12

t = 1
`(8)/(12)`

t = 1
`(2)/(3)`

there will be a max at 1.6666666666 seconds

-6*
`1.6666666666666^(2)` + 20 * 1.6666666666 +4

= 16.666666666666 + 33.333333333333 + 4

= - 16
`(2)/(3)` + 33
`(1)/(3)` + 4

= 20
`(2)/(3)` feet max height ( not too high, for a rocket)

time of flight:

0 = -6
`t^(2)` +20t + 4

use quadratic formula to find t

-20 +- sqrt [
`20^(2)` - 4*(-6)*4 ] / 2*(-6)

-20 +- sqrt [400 + 96 ] / -12

-20 +- sqrt [496 ] / -12

-20 +- 22.27105 / -12

try the negative option 1st

-42.27105 / -12

3.522 seconds. time of flight

when will the rocket be at 12' ? :

12 = -6
`t^(2)` +20t + 4

0 = -6
`t^(2)` +20t -8

use quadratic formula again to find t

-20 +- sqrt [
`20^(2)` - 4*(-6)*(-8) ] / 2*(-6)

-20 +- sqrt [ 400 - 192 ] / -12

-20 +- sqrt [208 ] / -12

-20 - 14.4222 / -12

-34.4222 / -12

2.8685 seconds ( on the way down)

and

-20 + 14.4222 / -12

-5.578 / - 12

0.4648 seconds ( on the way up )