Question

Can someone please solve these?

Answer 1

Answer:

Step-by-step 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 $\frac{8}{12}$

t = 1 $\frac{2}{3}$

there will be a max at  1.6666666666  seconds

-6*$1.6666666666666^{2}$  + 20 * 1.6666666666  +4

= 16.666666666666 + 33.333333333333 + 4

= - 16$\frac{2}{3}$ + 33 $\frac{1}{3}$ + 4

= 20$\frac{2}{3}$  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 )