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Answer :
The toy rocket is launched vertically from ground level, at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 72 m and acquired a velocity of 30 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance.
The total energy of the rocket, which is a sum of its kinetic energy and potential energy, is constant.
At a height of 72 m with the rocket moving at 30 m/s, the total energy is m*9.8*72 + (1/2)*m*30^2 where m is the mass of the rocket.
At ground level, the total energy is 0*m*9.8 + (1/2)*m*v^2.
Equating the two gives: m*9.8*72 + (1/2)*m*30^2 = 0*m*9.8 + (1/2)*m*v^2
=> 9.8*72 + (1/2)*30^2 = (1/2)*v^2
=> v^2 = 11556/5
=> v = 48.07
The velocity of the rocket when it impacts the ground is 48.07 m/s
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Final answer:
To find the speed of the rocket upon impact on the ground, we need to consider two phases: the burn phase and the unpowered flight phase. During the burn phase, we can calculate the acceleration using the equation v^2 = u^2 + 2as. In the unpowered flight phase, we find that the maximum height is zero and calculate the time taken for the rocket to reach the maximum height and fall back to the ground. Combining these results, we determine that the speed of the rocket upon impact on the ground is 4 times the initial velocity.
Explanation:
To find the speed of the rocket upon impact on the ground, we need to break down the motion of the rocket into two phases: the burn phase and the unpowered flight phase.
In the burn phase, the rocket rises to a height of 72m and acquires a velocity of 30 m/s. Since the rocket is launched vertically, the velocity acquired is purely vertical.
Using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement, we can find the acceleration during the burn phase:
30^2 = u^2 + 2 * a * 72
900 = u^2 + 144a
We also know that the acceleration during the burn phase is constant. So, we can say that a = g, where g is the acceleration due to gravity.
Substituting the value of a in our equation, we get:
900 = u^2 + 144g
We can rearrange the equation to solve for g:
g = (900 - u^2) / 144
Now, let's move on to the unpowered flight phase. In this phase, the rocket continues to rise until it reaches its maximum height, where its velocity becomes zero. At this point, the rocket starts falling back to the ground.
Since the rocket is in free fall, its acceleration is equal to the acceleration due to gravity, g.
Using the equation v^2 = u^2 + 2as again, we can find the maximum height reached by the rocket:
0 = v^2 + 2 * g * h_max
0 = 0 + 2 * g * h_max
h_max = 0
So, the maximum height reached by the rocket is 0m.
Now, we can find the time taken for the rocket to reach its maximum height during the unpowered flight phase using the equation v = u + gt:
0 = u + g * t_max
t_max = -u / g
Finally, we can find the time taken for the rocket to fall back to the ground using the equation s = ut + 0.5gt^2:
0 = u * t_fall + 0.5 * g * t_fall^2
t_fall = -2u / g
Since the rocket falls back to the ground, the time taken is positive.
Now, we can find the total time of flight:
t_total = t_burn + t_max + t_fall
t_total = 0s + (-u / g) + (-2u / g)
t_total = -3u / g
Since the time taken cannot be negative, we take the magnitude of the time:
t_total = 3u / g
Now, we can find the final velocity of the rocket upon impact on the ground using the equation v = u + gt:
v = u + g * t_total
v = u + g * (3u / g)
v = u + 3u
v = 4u
So, the speed of the rocket upon impact on the ground is 4 times the initial velocity.
Given that the initial velocity is 30 m/s, the speed of the rocket upon impact is 4 * 30 m/s = 120 m/s.
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