The height in feet of a bottle rocket in given by h(t)=160t-16t^2 where t is the time in seconds. How long will it take for the rocket to return to ground ? What is the height after 2 secs? HELP ME PLEASE

a) 10 sec

b) 256 feet

Step-by-step explanation:

a)

The height of the rocket above the ground at time t is described by the second-order equation

(1)

where:

t is the time, in seconds

h(t) is the height, in feet

represents the initial vertical velocity of the rocket

represents the acceleration due to gravity (downward)

The rocket will return to the ground when the height is zero, so when

Substituting into eq(1) and solving for t, we find:

which has two solutions:

t = 0 (initial instant, so when the rocket starts its motion)

t = 10 s --> this is the time at which the rocket returns to the ground

b)

The height of the rocket is given by

t is the time, in seconds

h(t) is the height, in feet

Here we want to find the height of the rocket after 2 seconds: to do that, we just need to substitute

t = 2 sec

into the equation of the height.

By doing so, we find:

Therefore, the height of the rocket after 2 seconds is 256 feet.

1/4 and 8/3 because:

the reciprocal if 4 is 1/4 because 4 is used as 4/1 and the opposite of that is 1/4. the opposite of 3/8 is 8/3 so that is the answer.

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,