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.
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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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
Thank you!
You are welcome.
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
To solve this problem, we will use the equations of motion for a vertically launched projectile.
First, let's calculate the velocity of the rocket when its engine runs out of fuel.
Using Newton's second law, we know that the net force acting on the rocket is equal to the product of its mass and acceleration.
F_net = m * a
The net force is the sum of the thrust force provided by the engine and the force due to gravity acting downward.
F_net = F_thrust - F_gravity
Since the rocket is starting from rest, its initial velocity is zero (v_initial = 0). The acceleration can be found using the equation:
a = (F_thrust - F_gravity) / m
Substituting the given values:
a = (25 N - m * g) / m
Let's calculate the acceleration:
a = (25 N - 1 kg * 9.8 m/s^2) / 1 kg
a = 15.2 m/s^2
Next, we can use the equation of motion for vertical acceleration to find the velocity of the rocket when the engine runs out of fuel:
v_final = v_initial + a * t
Substituting the known values:
v_final = 0 + 15.2 m/s^2 * 10 s
v_final = 152 m/s
a. To find the height above the ground when the engine runs out of fuel, we can use the equation for displacement:
s = v_initial * t + (1/2) * a * t^2
Since v_initial is 0, the equation simplifies to:
s = (1/2) * a * t^2
Substituting the known values:
s = (1/2) * 15.2 m/s^2 * (10 s)^2
s = 760 m
So, the rocket is 760 meters above the ground when the engine runs out of fuel.
b. After the engine runs out of fuel, the rocket will continue to climb until it comes to a stop. At the highest point, the vertical velocity will be zero. To find the maximum height, we can use the equation:
v_final = v_initial + a * t
Since we know v_final is 0, we can solve for t:
0 = 152 m/s - 9.8 m/s^2 * t
t = 152 m/s / 9.8 m/s^2
t = 15.51 s
To find the maximum height, we can use the equation for displacement:
s = v_initial * t + (1/2) * a * t^2
Substituting the known values:
s = 0 + (1/2) * 9.8 m/s^2 * (15.51 s)^2
s = 1,198.22 m
So, the rocket reaches a height of 1,198.22 meters above the point where the engine runs out of fuel.
c. To find the total time of flight, we need to consider the time it takes for the rocket to descend back to the ground. When the parachute deploys, the upward force (F_parachute) balances the force of gravity (F_gravity):
F_parachute = F_gravity
Substituting the known value for F_parachute:
5 N = m * g
Since the mass of the parachute is negligible, we can use this equation to find the time it takes for the rocket to descend back to the ground:
t_total = 2 * t_max_height
Substituting the known values:
t_total = 2 * 15.51 s
t_total = 31.02 s
Therefore, the total time of flight from launch to landing is 31.02 seconds.