46)
Since air resistance is ignored, both cannonballs will fall with the same acceleration due to gravity, which is approximately 9.8 m/s^2. The horizontal velocity of cannonball A does not affect its vertical motion, so it will fall at the same rate as cannonball B.
The time it takes for an object to fall to the ground from a certain height is given by the formula t = sqrt(2h/g), where t is the time, h is the initial height, and g is the acceleration due to gravity.
For cannonball B, which is dropped from a height of 20 meters, the time it takes to reach the ground is:
t = sqrt(2h/g) = sqrt(2*20/9.8) = 2.02 seconds
For cannonball A, which is fired horizontally with a velocity of 5 m/s, the time it takes to reach the ground is also 2.02 seconds, since the vertical motion is the same as cannonball B.
Therefore, both cannonballs will hit the ground at the same time.
47)
The recoil speed of the cannon can be calculated using the principle of conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it.
Before firing, the total momentum of the system is zero, since the cannon and cannonball A are at rest. After firing, the cannonball A has a momentum of 5 kg * 5 m/s = 25 kg m/s to the right, so the cannon must have an equal and opposite momentum to the left in order to conserve momentum.
The mass of the cannon is 500 kg, so its momentum after firing will be:
p = -25 kg m/s
The velocity of the cannon can be found using the equation:
p = mv
where p is the momentum, m is the mass, and v is the velocity. Solving for v, we get:
v = p/m = (-25 kg m/s) / (500 kg) = -0.05 m/s
Since the momentum of the cannon is negative, the recoil velocity is also negative, indicating that the cannon will move to the left after firing. The magnitude of the recoil velocity is 0.05 m/s, or approximately 0.18 km/h.