A girl throws a water-filled balloon at an angle of 50.0 degree(s) above the horizontal with a speed of 12.0 m/s. The horizontal component of the balloon's velocity is directed toward a car that is approaching the girl at a constant speed of 8.00 m/s. (See figure) (a) If the balloon is to hit the car at the same height at which it leaves her hand, what is the maximum distance the car can be from the girl when the balloon is thrown? You can ignore air resistance.

The car can intercept the balloon when the baloon is either on the way up or on the way down. You are interersted in the latter case, which gives the maximum distance of the car when thrown. The time it takes for the balloon to return to the same height that it was released is:

2*Vo*sin50/g = T = 1.876 s
The distance the car moves while the balloon is above the "throw" height is
8.00*T = 15.0 m

To find the maximum distance the car can be from the girl when the balloon is thrown, we need to consider the motion of the balloon vertically and horizontally.

Let's break down the motion into horizontal and vertical components.

Vertical motion:
The initial velocity (Viy) of the balloon in the vertical direction is given by:
Viy = V * sin(θ)
where V is the initial speed of the balloon (12.0 m/s) and θ is the launch angle (50.0 degrees).

Horizontal motion:
The initial velocity (Vix) of the balloon in the horizontal direction is given by:
Vix = V * cos(θ)
where V is the initial speed of the balloon (12.0 m/s) and θ is the launch angle (50.0 degrees).

Now, let's consider the time it takes for the balloon to hit the car. Since the car is approaching the girl at a constant speed, the horizontal distance traveled by the car will be the same as the horizontal distance traveled by the balloon.

The time taken for the balloon to hit the car can be found using the vertical motion equation:
y = Viy * t - (1/2) * g * t^2
where y is the vertical distance traveled, Viy is the initial vertical velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken.

Since the balloon hits the car at the same height it was thrown, the vertical distance traveled is zero:
0 = Viy * t - (1/2) * g * t^2

Solving this quadratic equation for t, we get:
t = (2 * Viy) / g

Now, we can calculate the horizontal distance traveled by the balloon (and the car):
x = Vix * t
where x is the horizontal distance traveled.

Plugging in the values:
Vix = V * cos(θ) = 12.0 m/s * cos(50.0 degrees)
Viy = V * sin(θ) = 12.0 m/s * sin(50.0 degrees)
g = 9.8 m/s^2

Using the derived equation for t:
t = (2 * Viy) / g = (2 * 12.0 m/s * sin(50.0 degrees)) / 9.8 m/s^2

Using the equation for x:
x = Vix * t = (12.0 m/s * cos(50.0 degrees)) * (2 * 12.0 m/s * sin(50.0 degrees)) / 9.8 m/s^2

Now, calculate the value of x to find the maximum distance the car can be from the girl when the balloon is thrown.

To find the maximum distance the car can be from the girl when the balloon is thrown, we need to consider the vertical and horizontal motion separately.

Let's start by analyzing the vertical motion of the water-filled balloon. We can use the kinematic equation:

y = yo + voy*t - 1/2*g*t^2

where:
y is the vertical displacement,
yo is the initial vertical position,
voy is the initial vertical velocity,
g is the acceleration due to gravity (approximately 9.8 m/s^2), and
t is the time.

Since the balloon is thrown at an angle of 50.0 degrees above the horizontal, the initial vertical velocity, voy, is given by:

voy = v * sin(θ)

where:
v is the initial velocity (12.0 m/s in this case),
θ is the launch angle (50.0 degrees).

We can determine the time it takes for the balloon to hit the car by setting y equal to 0 (since the balloon is thrown and caught at the same height) and solving for t:

0 = yo + voy*t - 1/2*g*t^2

0 = 0 + v * sin(θ) * t - 1/2*g*t^2

Using the quadratic formula, we can solve for t:

0 = -1/2*g*t^2 + v * sin(θ) * t

This equation has two solutions, t = 0 and t = (2 * v * sin(θ)) / g. Since the balloon is released and caught at the same height, we can discard the t = 0 solution. Therefore, the time it takes for the balloon to hit the car is:

t = (2 * v * sin(θ)) / g

Next, we need to consider the horizontal motion of the balloon. The horizontal distance traveled by the car during the time the balloon is in the air is given by:

x = vx * t

where:
x is the horizontal distance traveled by the car,
vx is the horizontal component of the car's velocity,
t is the time calculated earlier.

The horizontal component of the car's velocity, vx, is given as 8.00 m/s.

Finally, we can plug in the values into the equation to find the maximum distance the car can be from the girl when the balloon is thrown:

x = vx * t

x = 8.00 m/s * (2 * 12.0 m/s * sin(50.0 degrees) / 9.8 m/s^2)

Calculating this expression gives us the maximum distance the car can be from the girl when the balloon is thrown.