Answer: Assuming that the missile is launched vertically upward and neglecting air resistance, we can use conservation of energy to determine the missile's velocity at the peak of its trajectory. At the launch point, the missile has kinetic energy due to its speed and potential energy due to its height above the ground. At the peak of its trajectory, the missile has no kinetic energy (since it momentarily comes to a stop) but has maximum potential energy.
The total energy of the missile at any point in its trajectory is given by the sum of its kinetic and potential energies:
E = 1/2 mv^2 + mgh
where m is the mass of the missile, v is its velocity, g is the acceleration due to gravity, and h is its height above the ground.
At the launch point, the missile has a height of zero and a velocity of 6.0 km/s, so its total energy is:
E_launch = 1/2 m (6.0 km/s)^2 + mgh = 1/2 m (6.0 km/s)^2
At the peak of its trajectory, the missile has a height of 1000 km and a velocity of zero, so its total energy is:
E_peak = mgh = mg (1000 km)
Since energy is conserved, the total energy of the missile at the launch point must be equal to its total energy at the peak of its trajectory:
E_launch = E_peak
1/2 m (6.0 km/s)^2 = mg (1000 km)
Solving for v, the velocity at the peak of the trajectory, we get:
v = sqrt(2gh) = sqrt(2*9.81 m/s^2 * 1000 km) = 7.91 km/s
Therefore, the missile is moving at a speed of 7.91 km/s at the peak of its trajectory.
Explanation: