A model rocket has a mass of
3.00 kg. It is fired so that when
it is 220 m above the ground it is
travelling vertically upward at
165 m/s. At that point its fuel runs
out so that the rest of its flight is
without power. Assume that the
effect of air friction is negligible
and that all potential energies are
measured from the ground.
(a) What is the mechanical energy
of the rocket, relative to the
ground, when it is 220 m above
the ground?
(b) When it reaches the highest point
on its trajectory, what will its
gravitational potential energy be?
(c) How far above the ground is the
rocket at its highest point?
(d) When it hits the ground, what
is its speed?
To answer these questions, we need to consider the different forms of energy involved in the rocket's motion: kinetic energy (KE), gravitational potential energy (PE), and mechanical energy (ME). We can use the following equations:
1. Kinetic energy: KE = (1/2) * mass * velocity^2
2. Gravitational potential energy: PE = mass * gravitational acceleration * height
3. Mechanical energy: ME = KE + PE
Now let's go through each part of the question.
(a) What is the mechanical energy of the rocket, relative to the ground, when it is 220 m above the ground?
To find the mechanical energy, we need to calculate the kinetic energy and gravitational potential energy at that point. Since the rocket has only vertical motion, its kinetic energy is given by:
KE = (1/2) * mass * velocity^2 = (1/2) * 3.00 kg * (165 m/s)^2
(b) When it reaches the highest point on its trajectory, what will its gravitational potential energy be?
At the highest point in its trajectory, the rocket will have zero velocity, so its kinetic energy will be zero. Therefore, the mechanical energy of the rocket at the highest point will be entirely in the form of gravitational potential energy:
PE = mass * gravitational acceleration * height = 3.00 kg * 9.8 m/s^2 * 220 m
(c) How far above the ground is the rocket at its highest point?
The highest point of the rocket's trajectory occurs when its velocity becomes zero. This happens if we consider the upwards direction as positive. Using the equation:
vf = vi + at
where 'vf' is the final velocity, 'vi' is the initial velocity, 'a' is the acceleration, and 't' is the time, we have:
0 = 165 m/s - 9.8 m/s^2 * t
Solving for 't', we get:
t = 165 m/s / 9.8 m/s^2
To find the height at the highest point, we use the equation:
height = vi * t + (1/2) * a * t^2
Since the rocket is moving upwards, we consider all initial quantities positive:
height = 165 m/s * (165 m/s / 9.8 m/s^2) + (1/2) * (-9.8 m/s^2) * (165 m/s / 9.8 m/s^2)^2
(d) When it hits the ground, what is its speed?
When the rocket hits the ground, all of its mechanical energy will be converted to kinetic energy. Therefore, we can equate the initial mechanical energy to the final kinetic energy:
ME = KE
Solving for the final velocity 'v', we have:
KE = (1/2) * mass * v^2
Comparing this equation with the mechanical energy equation, we equate the right-hand sides:
(1/2) * 3.00 kg * (165 m/s)^2 = (1/2) * 3.00 kg * v^2
Solving for 'v', we find the rocket's speed when it hits the ground.
(a) To find the mechanical energy of the rocket when it is 220 m above the ground, we need to consider its kinetic energy and gravitational potential energy.
The kinetic energy (KE) of the rocket is given by the formula:
KE = (1/2) * mass * velocity^2
Substituting the values:
KE = (1/2) * 3.00 kg * (165 m/s)^2
KE = 1/2 * 3.00 kg * (27,225 m^2/s^2)
KE = 20,443.75 J
The gravitational potential energy (PE) can be calculated using the formula:
PE = mass * gravitational acceleration * height
Substituting the values:
PE = 3.00 kg * 9.8 m/s^2 * 220 m
PE = 6,468 J
The mechanical energy (ME) of the rocket is the sum of its kinetic and potential energies:
ME = KE + PE
ME = 20,443.75 J + 6,468 J
ME = 26,911.75 J
Therefore, the mechanical energy of the rocket, relative to the ground, when it is 220 m above the ground is 26,911.75 J.
(b) At the highest point of its trajectory, the rocket's velocity is zero. Therefore, its kinetic energy at that point is zero.
The gravitational potential energy (PE) at the highest point can be found using the same formula as before:
PE = mass * gravitational acceleration * height
Since the height is now maximum, the gravitational potential energy is also maximum.
Substituting the values:
PE = 3.00 kg * 9.8 m/s^2 * max height
To find the gravitational potential energy at the highest point, we need to find the maximum height the rocket reaches.
(c) To find the maximum height, we can use the conservation of mechanical energy principle. The initial mechanical energy (ME) of the rocket is equal to the mechanical energy at the highest point, neglecting air friction.
Initial ME (at 220 m above the ground) = ME at the highest point
Therefore, we can calculate the maximum height by rearranging the formula for gravitational potential energy:
Max height = PE / (mass * gravitational acceleration)
Using the values of PE and substituting them into the equation:
Max height = 6,468 J / (3.00 kg * 9.8 m/s^2)
Max height = 221.83 m
Therefore, the rocket is approximately 221.83 m above the ground at its highest point.
(d) When the rocket hits the ground, its potential energy is zero because it's at the ground level. Therefore, all of its initial mechanical energy must have transformed into kinetic energy.
Initially, the mechanical energy of the rocket (ME) was 26,911.75J, and at the ground, its potential energy is zero.
At the ground, the mechanical energy (ME) is equal to kinetic energy (KE):
ME = KE
Substituting the values:
26,911.75 J = (1/2) * mass * velocity^2
Rearranging the formula and solving for velocity:
velocity^2 = (2 * ME) / mass
velocity^2 = (2 * 26,911.75 J) / 3.00 kg
velocity^2 = 359,862.50 J / 3.00 kg
velocity^2 = 119,954.17 m^2/s^2
Taking the square root of both sides to find velocity:
velocity = √(119,954.17 m^2/s^2)
velocity ≈ 346.40 m/s
Therefore, the speed of the rocket when it hits the ground is approximately 346.40 m/s.