A basketball player is running at a constant speed of 2.5 m/s when a player tosses a basketball upward with a speed of 6.0 m/s. How far does the player run before he catches the ball? Ignore air resistance.

Options

A. 3.1 m
B. 4.5 m
C. The ball cannot be caught because it will fall behind the player.
D. 6.0 m

Answer

To solve the problem step by step, we can break it down into the following parts:

### Step 1: Understand the motion of the basketball
The basketball is thrown upward with an initial velocity (u) of 6.0 m/s. We need to calculate how long it will take for the ball to reach its highest point and then return to the same height (the time of flight).

### Step 2: Calculate the time of flight
The time of flight (T) for an object thrown vertically upwards can be calculated using the formula:
\[ T = \frac{2u}{g} \]
where:
- \( u = 6.0 \, \text{m/s} \) (initial velocity of the ball)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ T = \frac{2 \times 6.0}{9.8} = \frac{12.0}{9.8} \approx 1.2245 \, \text{s} \]

### Step 3: Calculate the distance the basketball player runs
The basketball player is running at a constant speed of 2.5 m/s. To find out how far the player runs while the ball is in the air, we can use the formula:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]

Substituting the values:
\[ \text{Distance} = 2.5 \, \text{m/s} \times 1.2245 \, \text{s} \approx 3.06125 \, \text{m} \]

### Step 4: Round the distance to a reasonable number
We can round this value to two decimal places:
\[ \text{Distance} \approx 3.06 \, \text{m} \]

### Conclusion
The basketball player runs approximately 3.06 meters before he catches the ball.

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