If y=(x^2-4)^5(3x+4)^4
1 answer:
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Answer:
Step-by-step explanation:
h = -6 +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
t = 1
there will be a max at 1.6666666666 seconds
-6* + 20 * 1.6666666666 +4
= 16.666666666666 + 33.333333333333 + 4
= - 16 + 33
+ 4
= 20 feet max height ( not too high, for a rocket)
time of flight:
0 = -6 +20t + 4
use quadratic formula to find t
-20 +- sqrt [ - 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 +20t + 4
0 = -6 +20t -8
use quadratic formula again to find t
-20 +- sqrt [ - 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 )