In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 3.7 m/s at an angle of 15.4° below the horizontal. It is released 0.83 m above the floor. What horizontal distance does the ball cover before bouncing?
m
To determine the horizontal distance the ball covers before bouncing, we need to use the equations of motion for projectile motion.
First, let's break down the given information:
- Initial speed (v₀) = 3.7 m/s
- Launch angle (θ) = 15.4° below the horizontal
- Initial height (h) = 0.83 m
We can start by calculating the time it takes for the ball to reach its peak height, which is given by the equation:
t_peak = v₀ * sin(θ) / g
Using the value for g (acceleration due to gravity) of approximately 9.8 m/s², we can now calculate the time to reach the peak:
t_peak = 3.7 * sin(15.4°) / 9.8
Next, we need to find the total time of flight (t_total) for the ball. Since the ball reaches its peak twice during its flight, we can use the formula:
t_total = 2 * t_peak
Substituting the value of t_peak, we can find t_total.
Now that we have the total time of flight, we can determine the horizontal distance (d) covered by the ball using the equation:
d = v₀ * cos(θ) * t_total
Substituting the values, we can calculate d.
Therefore, the horizontal distance the ball covers before bouncing is approximately __ meters.