Question

I need some help

Answer 1

Can you find dH/dt ?

Answer 2

yes
dh/dt = 128 + 16t -12t^2

Answer 3

then ?

Answer 4

dH/dt represents the change in the hight of the rocket. So if it is positive, means rocket is flying up. When it becomes negative means rocket start to go down

Answer 5

What moment in time is the velocity maximum?

Answer 6

find the zeros of dh/dt , t=4 ,t=-8/3
then we have to compare the velocities of t=0 , t=4 , t=8/3
v(0) =128 , v(4)= 0 minimum , v(8/3) =256/3
so at t=0 the velocity is maximum ? but its says wrong !

Answer 7

so you tried dH(0)/dt = v(0) =128 ft/sec

and it marked it as wrong?

Answer 8

yes

Answer 9

try 400/3

Answer 10

that's right but how did you get it ?

Answer 11

height = H
velocity = dH/dt

Answer 12

maximize velocity

Answer 13

solve : d/dt (velocity) = 0

Answer 14

t= 4 and t=-8/3 ?

Answer 15

H = 128t + 8t^2 - 4t^3
V = 128 + 16t - 12t^2

to maximize V, find its derivative and set it equal to 0

Answer 16

V' = 0

=>

16-24t = 0

Answer 17

that gives t = 2/3 as the only critical point

Answer 18

evaluate V at 2/3 :

V(t) = 128 + 16t - 12t^2
V(2/3) = ?

Answer 19

v'(t) = 16 -24t =0
t= 2/3
v(2/3) = 400/3

Answer 20

yes that means when the acceleration =0 the velocity will be the maximum

Answer 21

maybe put it like this : when the acceleration = 0, the velocity is NOT changing

Answer 22

ok
now (c) how can i find the max & min V over the first second ?

Answer 23

evaluate V at :

0, 1, 2/3

Answer 24

v(0) =128 minimum
v(1) =132
v(2/3) =400/3 maximum

Answer 25

thank you , you helped me a lot :)