Arocket rises vertically, from rest, with an acceleration of 3.2 m/s^2 until it runs out of fuel at an altitude of 725 m . after this point, its acceleration is that of gravity, downward. a)what is the velocity of the rocket when it runs out of fuel?
b)how long does it take to reach this point?
c)what maximum altitude does the rocket reach?
d)how much time (total) does it take to reach maximum altitude?
e)how much time (total) does it take to reach maximum altitude?
f)how long (total) is it in the air?

a) The velocity of the rocket when it runs out of fuel is 67 m/s

b) It takes the rocket 21 s to reach an altitude of 725 m

c) The maximum altitude of the rocket is 954 m

d) and e) It takes the rocket 28 s to reach the maximum altitude.

f) The rocket is 42 s in the air.

Explanation:

The equations for the height of the rocket are the following:

y = y0 + v0 · t + 1/2 · a · t²

and, when the rocket runs out of fuel:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the rocket at time t

y0 = initial height

v0 = initial velocity

a = acceleration due to engine of the rocket.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

t = time

Let´s place the origin of the frame of reference at the launching point.

b) First, let´s calculate the time it takes the rocket to reach an altitude of 725 m:

y = y0 + v0 · t + 1/2 · a · t²           (y0 = 0, v0 = 0)

725 m = 1/2 · 3.2 m/s² · t²

725 m / 1/2 · 3.2 m/s² = t²

t = 21 s

a)The velocity of the rocket is given by the following equation:

v = v0 + a · t     (v0 = 0)

v = 3.2 m/s² · 21 s = 67 m/s

c) After the rocket runs out of fuel, it starts slowing down until it comes to stop and starts to fall. Let´s calculate the time it takes the rocket to stop after running out of fuel. At that time, the rocket will be at its maximum height.

v = v0 + g · t

0 m/s = 67 m/s - 9.8 m/s² · t

-67 m/s / -9.8 m/s² = t

t = 6.8 s

The altitude reached after running out of fuel will be:

y = y0 + v0 · t + 1/2 · g · t²

y = 725 m + 67 m/s · 6.8 s - 1/2 · 9.8 m/s² · (6.8 s)²

y = 954 m

d) and e)

The maximum altitude will be reached after (21 s + 6.8 s) 28 s

f) Now, let´s calculate how much time it takes the rocket to return to the ground (y = 0) from the max-altitude:

y = y0 + v0 · t + 1/2 · g · t²

0 m = 954 m + 0 m/s · t - 1/2 · 9.8 m/s² ·t²

-954/- 1/2 · 9.8 m/s² = t²

t = 14 s

The rocket was (28 s + 14 s) 42 s in the air.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y