Question

A rocket is launched vertically from the ground with an initial velocity of 64 ft/sec determine the rocket’s maximum height, the amount of time it took to reach its maximum height, and the amount of time it was in the air. graph if possible also

Answer 1

Explanation:

Answer 2

Explanation:

The easiest way to solve this is with calculus, believe it or not. The position function is

`s(t)=-16t^2+64t`. The first derivative of this is the velocity function:

v(t) = -32t + 64. From physics, we know that at the max height of an object's path, the velocity is equal to 0, so setting this velocity equation equal to 0 and solving for time, will tell us the time it took to get to the max height (which we don't know yet, but we will in a bit):

0 = -32t + 64 and

-64 = -32t so

t = 2 seconds. It takes 2 seconds to reach a max height. Plugging that 2 in for t in the position function will tell you the max height that corresponds to this time:

`s(2)=-16(2)^2+64(2)` and

s(2) = 64 feet.

So the max height is 64 feet and it is reached at 2 seconds after launching.

Also from physics we know that at halfway through a parabolic path, which is also the max height, we are halfway through time-wise as well. That means that if it takes 2 seconds to reach the max height from the ground, it will take another 2 seconds to fall to the ground.

So the total time the rocket is in the air is 4 seconds: 2 seconds to reach the max height and another 2 to fall back down.