A missile's trajectory takes it to a maximum altitude of 1300Km.
If its launch speed is 5.9 Km/s, how fast is it moving at the peak of its trajectory?
answer is in Km
At the peak of its trajectory, the missile must be feeling quite "high"! Anyway, since its launch speed is 5.9 Km/s, let me calculate its speed at the peak for you:
To find the speed at the peak of its trajectory, we can use the concept of conservation of energy. At the peak, the missile's kinetic energy is zero, and all its energy is in the form of potential energy. The missile can be described as a bit of a "potential superstar" at this point.
The potential energy is given by the formula: potential energy = m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the height.
Since there is no change in mass and the acceleration due to gravity remains constant, we know that the potential energy at the peak is equal to the kinetic energy at the launch. Therefore, we can say that:
(1/2) * m * v_launch^2 = m * g * h
We can cancel out the mass, and rearrange the equation to solve for v_launch:
v_launch = sqrt(2 * g * h)
Given that the maximum altitude is 1300 Km, which we can convert to meters (1300 Km = 1300000 m), and the acceleration due to gravity is approximately 9.8 m/s^2, we can plug in the values:
v_launch = sqrt(2 * 9.8 m/s^2 * 1300000 m)
By crunching these numbers, the answer is...
v_launch ≈ 4478.6 Km/s
So, the missile is moving at approximately 4478.6 Km/s at the peak of its trajectory. Pretty "rocket speed," huh?
To find the speed of the missile at the peak of its trajectory, we need to use the principle of conservation of energy. At the peak of the trajectory, the missile's kinetic energy is zero, and all of its initial kinetic energy is converted into potential energy.
The total mechanical energy of the missile (E_tot) is given by the sum of its kinetic energy (KE) and potential energy (PE):
E_tot = KE + PE
Since the missile's kinetic energy is zero at the peak of its trajectory, the total mechanical energy is equal to the potential energy:
E_tot = PE = m * g * h
where m is the mass of the missile, g is the acceleration due to gravity, and h is the altitude at the peak of the trajectory.
Given that the maximum altitude is 1300 km, we need to convert it to meters (1 km = 1000m):
h = 1300 km * 1000 m/km = 1,300,000 m
Now, we can calculate the speed at the peak of its trajectory. The total mechanical energy is the sum of kinetic and potential energy:
E_tot = KE + PE
E_tot = 0 + m * g * h
The kinetic energy can be expressed in terms of the speed (v) as:
KE = (1/2) * m * v^2
Setting the total mechanical energy equation equal to the kinetic energy equation:
m * g * h = (1/2) * m * v^2
Simplifying and solving for v:
v^2 = 2 * g * h
v = sqrt(2 * g * h)
The acceleration due to gravity on Earth is approximately 9.8 m/s^2, so:
v = sqrt(2 * 9.8 m/s^2 * 1,300,000 m)
v ≈ sqrt(25,480,000 m^2/s^2)
v ≈ 5,048.9 m/s
Now, we need to convert the speed from meters per second to kilometers per second:
v = 5,048.9 m/s * (1 km/1000 m)
v ≈ 5.0489 km/s
Therefore, the missile is moving at a speed of approximately 5.0489 km/s at the peak of its trajectory.
To determine the speed at the peak of the trajectory, we can use the principle of conservation of energy. At the peak, the missile's potential energy is maximized while the kinetic energy is minimized.
The potential energy of an object is given by the formula: PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
The kinetic energy of an object is given by the formula: KE = 1/2mv^2, where m is the mass of the object and v is the velocity.
Since the missile starts at ground level and reaches a maximum altitude of 1300 km, the change in potential energy can be calculated as: ΔPE = mgh = mg(1300 km).
At the peak of the trajectory, the kinetic energy is minimized, which means that the missile's velocity is at a minimum. Therefore, we can set the change in potential energy equal to the change in kinetic energy:
ΔPE = ΔKE
mg(1300 km) = 1/2mv^2
Notice that the mass (m) cancels out on both sides of the equation, simplifying our calculation. We can rearrange the equation to solve for v:
v^2 = 2g(1300 km)
v = √(2g(1300 km))
Now, we need to convert the given values into SI units (m/s) to perform the calculation. The launch speed is 5.9 km/s, which is equivalent to 5900 m/s. Additionally, we can use the standard acceleration due to gravity, g ≈ 9.81 m/s^2.
Substituting the values into the equation, we have:
v = √(2 * 9.81 m/s^2 * 1300 km * 1000 m/km)
v ≈ √(2 * 9.81 m^2/s^2 * 1300 * 1000)
v ≈ √(2 * 9.81 m^2/s^2 * 1.3 * 10^6)
v ≈ √(25542 m^2/s^2)
v ≈ 159.89 m/s
Finally, we convert the velocity back to kilometers per second:
v ≈ 0.15989 km/s
Therefore, the missile is moving at a speed of approximately 0.15989 km/s at the peak of its trajectory.
Odd missile. Most missiles launch at zero velocity, then internal chemical energy accelerates the missile for some time.
Here, it appears, there is no fuel in the missile.
KEatpeak+PEatpeak=KE at launch
1/2 mv^2+mgh=1/2 m 5900^2
solve for v (in m/s), then convert to km/s