A rocket rises vertically, from rest, with an acceleration of 3.99 m/s2 until it runs out of fuel at an altitude of 775 m. After this point, its acceleration is due to gravity downwards. What is the speed of the rocket, in m/s, when it runs out of fuel?

Vf = 78.64 m/s

Explanation:

The rocket is travelling upward at a constant acceleration of 3.99 m/s² until it runs out of fuel. So, in order to calculate its velocity at the point, where it runs out of fuel, we can simply use 3rd equation of motion:

2as = Vf² - Vi²

where,

a = acceleration = 3.99 m/s²

s = distance or height covered by rocket till fuel runs out = 775 m

Vf = Final Velocity = ?

Vi = Initial velocity = 0 m/s   (Since, rocket starts from rest)

Therefore,

2(3.99 m/s²)(775 m) = Vf² - (0 m/s)²

Vf = √(6184.5 m²/s²)

Vf = 78.64 m/s

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not too sure thinking all accelerations are pointed down