Question

PLEASE YOU CAN ANSWER NUMBER 1 OR 2 OR BOTH PLEASE I JUST NEED HELP WITH BOTH AND NO ONE WILL ANSWER MY QUESTIONS!!!!!!!! :(

Answer 1

Answer:

Step-by-step explanation:

Given that a rocket path is given by

$h(x) = -4x^2 + 16x$

When it hits the ground h(x) =0

Hence we get

$-4x^2 + 16x=0\\x=0, 4$

i.e. after 4 minutes the rocket hits the ground

2) Given that

$s(t) = -16t^2 + 48t + 280.$

We have to find time t when height >280

$s(t) = -16t^2 + 48t + 280>280\\\\t=0, 3$

Between 0 and 3 seconds the object would be above the building.

Answer 2

1.) You plug a number into the variable x. Then you will solve the equation for h(x) to find the time it takes for the rocket to travel the path given.