To solve this problem, we need to analyze the motion of the table tennis ball thrown upwards at an angle with the vertical while also considering the horizontal force acting on it due to the wind. The goal is to find the angle \( \theta_0 \) such that the tangent of that angle, \( \tan \theta_0 \), corresponds to one of the given options.
Understanding the Forces at Play
The ball is thrown with an initial velocity \( u_0 = 10 \, \text{m/s} \) at an angle \( \theta_0 \) with respect to the vertical. This means we can break down the initial velocity into its vertical and horizontal components:
- Vertical component: \( u_{y} = u_0 \cdot \cos(\theta_0) \)
- Horizontal component: \( u_{x} = u_0 \cdot \sin(\theta_0) \)
Since the ball returns to the starting point, the time spent going up must equal the time spent coming down. The horizontal force acting on the ball is \( F = 0.08 \, \text{N} \), which will affect the horizontal motion.
Calculating the Time of Flight
The vertical motion is influenced by gravity, which we can denote as \( g = 9.81 \, \text{m/s}^2 \). The time \( t \) to reach the maximum height can be calculated using the formula:
At maximum height, the vertical velocity becomes zero:
0 = \( u_{y} - g \cdot t \)
Rearranging gives:
t = \( \frac{u_{y}}{g} = \frac{u_0 \cdot \cos(\theta_0)}{g} \)
The total time of flight \( T \) will be twice this time (up and down):
T = \( 2 \cdot \frac{u_0 \cdot \cos(\theta_0)}{g} \)
Horizontal Motion and Force
The horizontal motion is influenced by the wind force. The horizontal acceleration \( a_x \) due to the wind can be calculated using Newton's second law:
F = m \cdot a_x
Where \( m = 0.02 \, \text{kg} \) (since 20 g = 0.02 kg). Thus:
a_x = \( \frac{F}{m} = \frac{0.08}{0.02} = 4 \, \text{m/s}^2 \)
The horizontal distance \( d \) traveled during the time \( T \) is given by:
d = \( u_{x} \cdot T + \frac{1}{2} a_x \cdot T^2 \)
Setting Up the Equation
Since the ball returns to the starting point, the horizontal distance must equal zero:
0 = \( u_{x} \cdot T + \frac{1}{2} a_x \cdot T^2 \)
Substituting \( T \) and the components of \( u_0 \):
0 = \( (u_0 \cdot \sin(\theta_0)) \cdot \left(2 \cdot \frac{u_0 \cdot \cos(\theta_0)}{g}\right) + \frac{1}{2} \cdot 4 \cdot \left(2 \cdot \frac{u_0 \cdot \cos(\theta_0)}{g}\right)^2 \)
Solving for \( \tan \theta_0 \)
After simplifying and rearranging the equation, we can derive the relationship between \( \tan \theta_0 \) and the given forces. Ultimately, we find that:
\( \tan \theta_0 = \frac{a_x \cdot g}{u_0^2} \)
Substituting the known values:
\( \tan \theta_0 = \frac{4 \cdot 9.81}{10^2} = \frac{39.24}{100} = 0.3924 \)
Choosing the Correct Option
From the options provided:
- (A) 0.2
- (B) 0.4
- (C) 2.5
- (D) 1.2
The closest value to \( 0.3924 \) is \( 0.4 \). Therefore, the correct answer is:
(B) 0.4
