A small plane moving at 207km/hr flying 50.0m above the ground chases a convertible moving at 105km/hr down a straight highway. The plane has a package to drop in the car’s back seat. Assuming the vehicles maintain their speed and altitude and there is no air resistance, how far should the pilot be behind the car when he lets go of the package?

To determine the distance the pilot should be behind the car when he lets go of the package, we need to consider the relative velocity between the plane and the car.

Let's first convert the speeds of the plane and the car to meters per second (m/s) for consistent units:
Plane speed = 207 km/hr = (207 * 1000) / (60 * 60) m/s = 57.5 m/s
Car speed = 105 km/hr = (105 * 1000) / (60 * 60) m/s = 29.2 m/s

Since the plane and car are moving in the same direction, the relative velocity is the difference between their velocities. Thus, the relative velocity between the plane and the car is:
Relative velocity = Plane speed - Car speed = 57.5 m/s - 29.2 m/s = 28.3 m/s

Now, to find the time it takes for the package to reach the car, we need to consider the vertical component of the motion. The plane is flying at a constant altitude of 50.0 m above the ground, so the only vertical motion is the package dropping vertically.

The time it takes for the package to fall can be determined using the equation for vertical motion: h = (1/2) * g * t^2, where h is the height (50.0 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

Rearranging the equation, we have:
t^2 = (2h) / g
t = sqrt((2h) / g) = sqrt((2 * 50.0) / 9.8) = 3.19 s (approx.)

Now, using the time it takes for the package to fall, we can calculate the distance the car travels in that time. The distance is simply the product of the relative velocity and the time:
Distance = Relative velocity * Time = 28.3 m/s * 3.19 s = 90.24 m (approx.)

Therefore, the pilot should be approximately 90.24 meters behind the car when he lets go of the package.