A disgruntled physics student sees the front end of his teacher's car below him at the base of a building. If the student is 5.0 m high above the car which is 3.0 m long and traveling only 3m/s to the right, will the apple of the student hit the teacher's car? What is the initial velocity and direction of the apple?

We will assume the student was located at a point perpendicular to the direction of motion of the car.

We will restrict the answer to throwing the apple in a plane perpendicular to the motion of the car, thus a two-dimentional problem.
The apple has to hit somewhere on the car within the time the car is immediately facing the student, thus
t≤3m/(3m/s)=1 sec.

It is not clear if the student is right above the car because there is no lateral distance given, but question asks for direction.

Assuming student is right on top of the passing car, so the direction would be either up or down. Preferably down because there is only one second to hit it.

For a velocity u, through height h, and all measurements are positive upwards, we have
h ≥ ut+(1/2)(-9.8)t²
with t=1, the equation becomes
-5≥u(1)-(1/2)9.8(1²)
Solving for u gives u≤-0.1 m/s

To determine if the apple will hit the teacher's car, we need to analyze the motion of the apple and the car separately and then compare their positions.

1. Let's start by finding the time it takes for the apple to fall from a height of 5.0 m. We can use the equation for the vertical motion of an object in free fall:
h = (1/2)gt^2
where h is the vertical distance, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

Plugging in the values:
5.0 m = (1/2)(9.8 m/s^2)t^2
10.0 m = 4.9 m/s^2 * t^2
t^2 = (10.0 m) / (4.9 m/s^2)
t^2 ≈ 2.04 s^2
t ≈ √2.04 s
t ≈ 1.43 s

Therefore, it takes approximately 1.43 seconds for the apple to fall from a height of 5.0 m.

2. Now let's determine how far the car will travel during that time. We can use the equation for uniform motion:
s = v*t
where s is the distance, v is the velocity, and t is the time.

Plugging in the values:
s = (3.0 m/s)(1.43 s)
s ≈ 4.29 m

Therefore, the car will travel approximately 4.29 m to the right during the time it takes for the apple to fall.

3. Now we can compare the distance the apple falls to the horizontal distance traveled by the car. If the apple's vertical distance is greater than or equal to the car's horizontal distance, it will hit the car.

The apple's vertical distance is 5.0 m, and the car's horizontal distance is 4.29 m. Since 5.0 m is larger than 4.29 m, the apple will hit the car.

4. Finally, let's determine the initial velocity and direction of the apple. Since the apple is falling vertically, its initial velocity in the vertical direction is 0 m/s. However, since the apple is dropped from a height, its initial velocity in the horizontal direction is the same as the car's velocity, which is 3.0 m/s to the right.

Therefore, the initial velocity of the apple is 3.0 m/s to the right in the horizontal direction, and 0 m/s in the vertical direction.