A model rocket with mass 1 kg is launched from the ground, starting from rest, and accelerates vertically due to an upward thrust of 25N being produced by the rocket’s engine. The rocket accelerates vertically for a time of 10 seconds at which point the rocket’s engine runs out of fuel. Assume that the rocket’s engine runs out of fuel instantaneously and stops operating and that the fuel contributes a negligible amount to the rocket’s mass.

Question

a. At the point at which the rocket’s engine runs out of fuel, how high above the ground is the rocket and what is its vertical speed?
b. Once the rocket’s engine runs out of fuel, the rocket continues to climb against gravity until it comes to rest? To what height above the point at which the rocket runs out of fuel does the rocket reach?

c. When the rocket reaches its highest point above the ground, a parachute deploys instantaneously and the parachute produces an upward force of 𝐹 = 5𝑁 to
slow the rate of decent of the rocket back to the ground. What is the total time of flight of the projectile from its launch to the time it lands back on the ground again? Assume that the mass of the parachute is negligible.

Thanks in advance!
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Answer 1

F = 25 - m g = 25 - 1*9.81 = 15.2 Newtons net force up
F = m a
a = 15.2 / 1 = 15.2 m/s^2
v = 0 + a t = 15.2 t
h = 0 + 0 t + (1/2) a t^2 = (15.2/2) t^2
at t = 10 seconds:
v = 152 m/s up
h = 7.6 (100) = 706 meters up at speed up =152 m/s
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now it coasts how much higher
v = 152 - g t = 152 - 9.81 t
top when v = 0
9.81 t = 152
t = 15.5 seconds more drifting up
additional h = Vi t - 4.9 t^2
= 152 * 15.5 - 4.9* (15.5)^2
= 2355 - 1177 = 1,177 meters MORE up
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now on the way down
Fdown = mg-5 = 9.81 - 5 = 4.81 Newtons down
a down = 4.81 / 1 = 4.81 m/s^2 (like about half of gravity)
1177 meters own = (1/2)(4.81) t^2
solve for t down
remember to add the ten seconds and the 15.5 seconds from the upward journey to get total time in sky

Answer 2

Thank you!

Answer 3

You are welcome.

Answer 4

I left out the initial 706 meters up during the fall to the ground
(1177 + 706 ) meters down = (1/2)(4.81) t^2
solve for t down