Arocket rises vertically, from rest, with an acceleration of 3.2 m/s2 until it runs out of fuel at an altitude of 950 m. after this point, its acceleration is that of gravity, down­ ward. (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) with what velocity does it strike the earth? (/) how long (total) is it in the air?

a) The velocity of the rocket when it runs out of fuel is 77 m/s.

b) It takes the rocket 24 s to reach the point where it runs out of fuel.

c) The maximum altitude reached is 1252 m.

d) It takes the rocket 32 s to reach its maximum height.

e) The rocket strikes the Earth with a velocity of -157 m/s.

f) In total, the rocket is 48 s in the air.

Explanation:

Hi there!

The height and velocity of the rocket can be calculated as follows:

y = y0 + v0 · t + 1/2 · a · t²     (when there is still fuel).

y = y0 + v0 · t + 1/2 · g · t²     ( after the rocket runs out of fuel)

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

a = upward acceleration due to the engines of the rocket.

g = downward acceleration due to gravity (-9.8 m/s² ).

In the same way, there will be two equations for velocity:

v = v0 + a · t  (before running out of fuel)

v = v0 + g · t (after running out of fuel)

Where "v" is the velocity at time "t".

a) When the rocket runs out of fuel, it is at an altitude of 950 m. Let´s find the time it takes the rocket to reach that height using the equation of height:

y = y0 + v0 · t + 1/2 · a · t²      (y0 and v0 = 0 because the rocket starts from rest and the origin of the frame of reference is located at the launching point on the ground)

950 m = 1/2 · 3.2 m/s² · t²

2 · 950 m / 3.2 m/s² = t²

t = 24 s

Now, with this time, we can calculate the velocity:

v = v0 + a · t

v = 0 m/s + 3.2 m/s² · 24 s

v = 77 m/s

b) It takes the rocket 24 s to reach the point where it runs out of fuel (calculated above).

c) After reaching 950 m, the acceleration of the rocket is downward. There is a moment at which the velocity is 0 and immediately after, the rocket starts to fall with negative velocity. In that instant, the rocket is at its maximum height.

Then, using the equation for velocity we can find at which time the rocket is at maximum height. Notice that the rocket was traveling with a velocity of 77 m/s due to the acceleration of the engines when it ran out of fuel and started to slow down due to gravity (v0 = 77 m/s):

v = v0 + g · t

0 = 77 m/s - 9.8 m/s² · t

-77 m/s / -9.8 m/s² = t

t = 7.9 s

It takes the rocket 7.9 s to reach its maximum height after it runs out of fuel.

Now, with this time and the equation of height, we can calculate the maximum height reached:

y = y0 + v0 · t + 1/2 · g · t²

y = 950 m + 77 m/s · 7.9 s - 1/2 · 9.8 m/s² · (7.9s)²

y = 1252 m

d) It takes the rocket 24 s to reach an altitude of 950m and 7.9 s more to reach its maximum height. In total 32 s.

e) First, let´s find how much time it takes the rocket to reach the ground (y = 0) from the maximum height (y0 = 1252 m and v0 = 0):

y = y0 + v0 · t + 1/2 · g · t²

0 = 1252 m +0 m/s · t -1/2 · 9.8 m/s² · t²

-1252 m / -4.9 m/s² = t²

t = 16 s

Now, with the equation of velocity:

v = v0 + g · t

v = 0 m/s -9.8 m/s² · 16 s

v = -157 m/s

f) The total time that the rocket is in the air is (24 s + 7.9 s + 16 s) 48 s.

Have a nice day!

tbh i dont know ether