A student performs an experiment to determine how the range of a ball depends on the velocity with which it is projected. The "range" is the distance between the points where the ball lends and from where it was projected, assuming it lands at the same height from which it was projected. It each trial, the student uses the same baseball, and launches it at the same angle. Table shows the experimental results.
`|{:("Trail","Launch speed" (m//s),"Range"(m)),(1,10,8),(2,20,31.8),(3,30,70.7),(4,40,122.5):}|`
Based on this data, the student then hypothesizes that the range, R, depends on the initial speed `v_(0)` according to the following equation : `R=Cv_(0)^(n)`, where C is a constant and n is another constant.
The student speculates that the constant C depends on :-
(i) The angle at which the ball was launched (ii) The ball's mass (iii) The ball's diameter
If we neglect air resistance, then C actually depends on :-

Options

A. I only
B. I and II
C. I and III
D. I, II and III

Explanation

To solve the problem step-by-step, we will analyze the data provided and derive the relationship between the range \( R \) and the initial speed \( v_0 \) of the ball.

### Step 1: Understand the relationship
The student hypothesizes that the range \( R \) depends on the initial speed \( v_0 \) according to the equation:
\[ R = C v_0^n \]
where \( C \) is a constant and \( n \) is another constant.

### Step 2: Use the experimental data
The experimental data provided is as follows:

| Trial | Launch Speed \( v_0 \) (m/s) | Range \( R \) (m) |
|-------|-------------------------------|--------------------|
| 1 | 10 | 8 |
| 2 | 20 | 31.8 |
| 3 | 30 | 70.7 |
| 4 | 40 | 122.5 |

### Step 3: Set up equations based on the data
Using the equation \( R = C v_0^n \), we can set up equations for each trial:

1. For Trial 1:
\[ 8 = C \cdot 10^n \]
2. For Trial 2:
\[ 31.8 = C \cdot 20^n \]
3. For Trial 3:
\[ 70.7 = C \cdot 30^n \]
4. For Trial 4:
\[ 122.5 = C \cdot 40^n \]

### Step 4: Divide equations to eliminate \( C \)
To find \( n \), we can divide the equations for Trials 2 and 1:
\[ \frac{31.8}{8} = \frac{C \cdot 20^n}{C \cdot 10^n} \implies \frac{31.8}{8} = \frac{20^n}{10^n} = 2^n \]
Calculating \( \frac{31.8}{8} \):
\[ \frac{31.8}{8} \approx 3.975 \implies 2^n \approx 3.975 \]

### Step 5: Solve for \( n \)
Taking logarithm base 2:
\[ n \approx \log_2(3.975) \approx 2 \]
Thus, we find that \( n = 2 \).

### Step 6: Substitute \( n \) back into the equation
Now we can rewrite the equation for range:
\[ R = C v_0^2 \]

### Step 7: Determine the constant \( C \)
Using any trial data to find \( C \). Let's use Trial 1:
\[ 8 = C \cdot 10^2 \implies C = \frac{8}{100} = 0.08 \]

### Step 8: Analyze the dependence of \( C \)
The student speculates that \( C \) depends on:
1. The angle at which the ball was launched
2. The ball's mass
3. The ball's diameter

However, if we neglect air resistance, the constant \( C \) actually depends primarily on the angle of projection \( \theta \) because the range formula in projectile motion is given by:
\[ R = \frac{v_0^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity. Thus, \( C \) is influenced by the angle \( \theta \).

### Final Answer
Therefore, the correct answer is that \( C \) actually depends on:
- (i) The angle at which the ball was launched.