Answer: 8.43 m
Explanation:
Given
Mass of the rocket, m = 600 g = 0.6 kg
Speed of riling, v = 2.5 m/s
F = mg, where
F = weight of the rocket in N
m = mass of the rocket in kg
g = acceleration due to gravity in m/s²
F = 0.6 * 9.8
F = 5.88 N
Since the the vertical thrust on the rocket is 8 N and 5.88 N of it will be used to counter the rockets weight. Thus, the remaining 2.12 N is what is available to provide acceleration.
Upward acceleration of the rocket is..... F = ma, a = F / m
a = 2.12 / 0.6
a = 3.53 m/s²
using equation of motion, we will calculate how long it will take to rise 20 m in the air
S = ut + 1/2at², where u = 0
20 = 1/2 * 3.53 * t²
t² = 4. / 3.53
t² = 11.33
t = √11.33 = 3.37 s
The distance then is
v = Δx / t, such that
Δx = v * t
Δx = 2.5 * 3.37
Δx = 8.43 m
Answer:
8.66 m
Explanation:
600g = 0.6 kg
Let g = 10m/s2. Gravity acting on the rocket model would have a magnitude of:
mg = 0.6*10 = 6 N
When the thrust of 8N acting on the rocket, the net force would be F = 8 - 6 = 2N vertically upward. So the net acceleration according to Newton 2nd law would be
a = F/m = 2 / 0.6 = 3.33 m/s2
We can calculate the time t it would take to reach h = 20m vertically at this acceleration:
[tex]h = at^2/2[/tex]
[tex]t^2 = 2h/a = 2*20/3.33 = 12[/tex]
[tex]t = \sqrt{12} = 3.46 s[/tex]
Assume that the horizontal speed v = 2.5m/s stays constant, then the horizontal distance from where it starts thrusting up to the loop is
2.5 * 3.46 = 8.66 m
