A girl throws a water-filled balloon at an angle of 50.0∘^\circ above the horizontal with a speed of 12.0m/s{\rm m}/{\rm s} . The horizontal component of the balloon's velocity is directed toward a car that is approaching the girl at a constant speed of 8.00m/s{\rm m}/{\rm s} .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.
Vo = 12m/s[50o]
Xo = 12*Cos50 = 7.71 m/s.
Yo = 12*sin50 = 9.19 m/s.
Tr = -Yo/g = -9.19/-9.8 = 0.938 s.
Tf = Tr = 0.938 s. = Fall time.
Tr+Tf = 0.938 + 0.938 = 1.876 s. = Time
to return to launching height.
Range=Xo * (Tr+Tf) = 7.71 * 1.876=14.46m
D max = 14.46 + 8m/s * 1.876s
D max = 14.46 + 15 = 29.5 m.
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To determine the maximum distance the car can be from the girl when the balloon is thrown, we need to consider the motion of the balloon and the car.
Let's break down the problem step-by-step:
Step 1: Find the initial vertical velocity of the balloon.
The given information states that the balloon is thrown at an angle of 50.0° above the horizontal with a speed of 12.0 m/s. To find the initial vertical velocity (Viy) of the balloon, we can use trigonometry.
Viy = V * sin(θ)
Viy = 12.0 m/s * sin(50.0°)
Viy ≈ 9.23 m/s
Step 2: Calculate the time it takes for the balloon to hit the car.
Since the balloon is thrown vertically, only the vertical motion needs to be considered. We will use the equation:
Δy = Viy * t + (1/2) * g * t^2
In this case, Δy is the height at which the balloon is thrown and t is the time it takes for the balloon to hit the car. Since we want the balloon to hit the car at the same height it is thrown, Δy will be zero.
0 = (9.23 m/s) * t + (1/2) * (-9.8 m/s^2) * t^2
This quadratic equation can be solved to find the time t. The solutions are t = 0 s and t ≈ 1.87 s. Since we are considering the time it takes for the balloon to hit the car, we choose the positive value, t = 1.87 s.
Step 3: Determine the horizontal distance traveled by the car in this time.
The horizontal distance traveled by the car can be calculated using the equation:
Δx = Vx * t
Where Δx is the horizontal distance, Vx is the horizontal component of the balloon's velocity, and t is the time it takes for the balloon to hit the car.
Since the balloon is thrown at an angle of 50.0° above the horizontal and the horizontal component of the speed is directed toward the car, we can find Vx:
Vx = V * cos(θ)
Vx = 12.0 m/s * cos(50.0°)
Vx ≈ 7.67 m/s
Δx = 7.67 m/s * 1.87 s
Δx ≈ 14.36 m
Therefore, the maximum distance the car can be from the girl when the balloon is thrown is approximately 14.36 meters.
To find the maximum distance the car can be from the girl when the balloon is thrown, we need to consider the horizontal motion and vertical motion of the balloon separately.
Let's first analyze the horizontal motion. The horizontal component of the balloon's velocity is directed toward the car, and the car is approaching the girl at a constant speed. This means that the horizontal distance the balloon travels is equal to the horizontal distance the car travels during the time of flight of the balloon.
Now, let's consider the vertical motion of the balloon. The balloon is thrown at an angle of 50.0° above the horizontal, so we can split the initial velocity of the balloon into its horizontal and vertical components.
The horizontal component of the initial velocity can be found using the formula:
Vx = V * cos(θ)
Where:
Vx = horizontal component of velocity
V = initial velocity of the balloon (12.0 m/s)
θ = angle of projection (50.0°)
Plugging in the values:
Vx = 12.0 m/s * cos(50.0°)
Vx = 12.0 m/s * 0.6428
Vx ≈ 7.714 m/s
The vertical component of the initial velocity can be found using the formula:
Vy = V * sin(θ)
Where:
Vy = vertical component of velocity
V = initial velocity of the balloon (12.0 m/s)
θ = angle of projection (50.0°)
Plugging in the values:
Vy = 12.0 m/s * sin(50.0°)
Vy = 12.0 m/s * 0.7660
Vy ≈ 9.192 m/s
Now, we can find the time of flight (t) of the balloon, which is the total time it takes for the balloon to reach its maximum height and then hit the car. The time of flight can be found using the vertical component of velocity and the acceleration due to gravity (g):
t = 2 * Vy / g
Where:
t = time of flight
Vy = vertical component of velocity (9.192 m/s)
g = acceleration due to gravity (9.8 m/s^2)
Plugging in the values:
t = 2 * 9.192 m/s / 9.8 m/s^2
t ≈ 1.869 s
Since the horizontal distance traveled by the balloon is equal to the horizontal distance traveled by the car during this time, we can calculate this distance using the horizontal component of velocity and the time of flight:
Distance = Vx * t
Where:
Distance = horizontal distance traveled by the balloon (and car)
Vx = horizontal component of velocity (7.714 m/s)
t = time of flight (1.869 s)
Plugging in the values:
Distance = 7.714 m/s * 1.869 s
Distance ≈ 14.432 m
Therefore, the maximum distance the car can be from the girl when the balloon is thrown is approximately 14.432 meters.