A volleyball is sent over a net, having been launched at a speed 4.6 m/s at an angle 28.2 degrees above the horizontal, from the ground. What is the maximum height, in m, that the net can be placed such that they ball can make it over?
y = (v sin28.2)^2/2g
To determine the maximum height of the net that the volleyball can clear, we need to analyze its projectile motion. We can break down the motion into horizontal and vertical components.
1. Horizontal Motion:
The initial velocity of the volleyball can be divided into its horizontal and vertical components using trigonometry. Since the launch angle is given as 28.2 degrees above the horizontal, we can find the horizontal component using the formula:
Horizontal component = Initial velocity × cos(angle)
In our case, the initial velocity is 4.6 m/s, so the horizontal component is calculated as follows:
Horizontal component = 4.6 m/s × cos(28.2°)
Horizontal component = 4.6 m/s × 0.882
Horizontal component ≈ 4.05 m/s
2. Vertical Motion:
The vertical component of the initial velocity can be determined in a similar fashion using trigonometry. We can calculate it using the formula:
Vertical component = Initial velocity × sin(angle)
In our case, the initial velocity is 4.6 m/s, so the vertical component is calculated as follows:
Vertical component = 4.6 m/s × sin(28.2°)
Vertical component = 4.6 m/s × 0.485
Vertical component ≈ 2.23 m/s
3. Maximum Height:
To calculate the maximum height, we need to find the time it takes for the volleyball to reach its peak. At the peak, the vertical velocity will be zero. You can use the following formula to calculate the time of flight:
Time of flight = (2 × vertical component) / (acceleration due to gravity)
The acceleration due to gravity is approximately 9.8 m/s². Plugging in the values:
Time of flight = (2 × 2.23 m/s) / 9.8 m/s²
Time of flight ≈ 0.46 s
Now that we have the time of flight, we can calculate the maximum height using the vertical component and the time of flight. We can use the formula:
Maximum height = vertical component × time of flight - (0.5 × acceleration due to gravity × time of flight²)
Plugging in the values:
Maximum height = 2.23 m/s × 0.46 s - (0.5 × 9.8 m/s² × (0.46 s)²)
Maximum height ≈ 0.52 m
Therefore, the maximum height the net can be placed so that the ball can clear it is approximately 0.52 meters.