a convertible travels down a straight, level road at a slow speed of 13 km/h. A person in the car throws a ball with a speed of 3.6 m/s forward at an angle of 30 degrees to the horizontal. Where is the car when the ball lands?

To determine where the car is when the ball lands, we first need to find the time it takes for the ball to land. We can use the equation of motion in the vertical direction for a projectile:

y = y0 + v0y * t - (1/2) * g * t^2

Where:
y = vertical displacement (in this case, the height above the ground)
y0 = initial vertical position (we can assume it is 0 since the car is on the ground)
v0y = vertical component of initial velocity
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time

We know the initial velocity (v0) of the ball is 3.6 m/s at an angle of 30 degrees to the horizontal, so we can find v0y:

v0y = v0 * sin(theta)

v0y = 3.6 m/s * sin(30 degrees)
v0y = 3.6 m/s * 0.5
v0y = 1.8 m/s

Next, we can find the time it takes for the ball to reach the ground by setting y = 0:

0 = 0 + 1.8 m/s * t - (1/2) * 9.8 m/s^2 * t^2

This is a quadratic equation, so we can use the quadratic formula:

t = (-v0y ± √(v0y^2 - 4 * (-(1/2) * g * y0)) / (2 * (-(1/2) * g))

Plugging in the values, we get:

t = (-1.8 m/s ± √(1.8^2 - 4 * (-(1/2) * 9.8 m/s^2 * 0)) / (2 * (-(1/2) * 9.8 m/s^2))

Simplifying further:

t = (-1.8 m/s ± √(3.24 m^2/s^2) / (-9.8 m/s^2)
t = (-1.8 m/s ± 1.8 m/s) / (-9.8 m/s^2)

Taking the positive value (as time cannot be negative), we have:

t = (1.8 m/s - 1.8 m/s) / (-9.8 m/s^2)
t = 0 / (-9.8 m/s^2)
t = 0 seconds

Since the time is 0 seconds, it means the ball lands immediately after it is thrown. Therefore, the car is at the same position when the ball lands.

To find where the car is when the ball lands, we need to calculate the horizontal distance traveled by the ball while it is in the air.

First, let's convert the speed of the car from km/h to m/s:
13 km/h = (13 * 1000) m / (60 * 60) s ≈ 3.61 m/s

Now, let's analyze the motion of the ball separately in the horizontal and vertical directions.

Horizontal motion:
The horizontal speed of the ball remains constant throughout the flight since there is no horizontal force acting on it. The horizontal speed is given by the equation:
Vx = V * cosθ,
where Vx is the horizontal component of velocity, V is the initial speed of the ball, and θ is the angle of projection.

Vx = 3.6 m/s * cos(30°) ≈ 3.6 m/s * 0.866 = 3.119 m/s.

Vertical motion:
The vertical motion of the ball is affected by the force of gravity. The vertical speed of the ball changes due to the acceleration due to gravity (g ≈ 9.8 m/s²). The time it takes for the ball to reach the ground can be found using the vertical equation of motion:
Δy = Vy * t - (1/2) * g * t²,
where Δy is the vertical distance covered, Vy is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.

Δy = 0 (since the ball lands on the ground)
Vy = V * sinθ,
g = 9.8 m/s².

Putting the values into the equation:
0 = (V * sinθ) * t - (1/2) * (9.8) * t²
0 = 3.6 m/s * sin(30°) * t - (4.9) * t²,
0 = 1.8 m/s * t - 4.9 m/s² * t².

Simplifying the equation, we get a quadratic equation:
4.9 t² - 1.8 t = 0,
t(4.9 t - 1.8) = 0.

To solve for t, we set each factor equal to zero:
t₁ = 0 (ignored since we need the time when the ball lands)
4.9 t - 1.8 = 0,
4.9 t = 1.8,
t = 1.8 / 4.9 ≈ 0.367 s.

Now that we have the time taken by the ball to reach the ground, we can calculate the horizontal distance traveled by the ball:
Δx = Vx * t.

Δx = 3.119 m/s * 0.367 s ≈ 1.146 m.

Therefore, when the ball lands, the car would have traveled approximately 1.146 meters horizontally down the road from its initial position.