A volleyball player spikes the ball from a height of 2.44 meters. Assume that the path of the ball is a straight line. To the nearest degree, what is the maximum angle, θ, at which the ball can be hit and land within the court?
You have to know the size of the court. Assuming the spike is driven parallel to the side lines, then if the distance to the far line is x, and assuming you want the angle from the vertical (otherwise you'd be asking for the minimum angle), then
x/2.44 = tan θ
See how important it is to completely specify the parameters of the problem? I may have assumed several things which were not correct. You can, of course, correct them and review your basic trig functions to complete the solution.
To find the maximum angle at which the ball can be hit and land within the court, we need to consider the vertical distance and the horizontal distance traveled by the ball.
Let's assume the horizontal distance traveled by the ball is x meters.
The vertical distance traveled by the ball can be calculated using the equation of motion for free fall: s = ut + (1/2)gt^2
Where:
s is the vertical distance (2.44 meters)
u is the initial vertical velocity (0 m/s, as the ball is initially at rest)
g is the acceleration due to gravity (-9.8 m/s^2)
t is the time taken (unknown)
Using the above equation, we can solve for time (t):
2.44 = 0 * t + (1/2)(-9.8)t^2
2.44 = -4.9t^2
t^2 = 2.44 / -4.9
t^2 = 0.4989
t ≈ √(0.4989)
t ≈ 0.706 seconds
Now, let's calculate the horizontal distance (x) traveled by the ball using the formula x = ut, where u is the horizontal velocity.
Since there is no horizontal acceleration, the horizontal velocity remains constant throughout the flight of the ball.
We can assume a typical horizontal velocity for a volleyball spike, which is around 18 m/s.
x = (18 m/s) * (0.706 seconds)
x ≈ 12.708 meters
Now we have the vertical distance (2.44 meters) and the horizontal distance (12.708 meters).
To find the maximum angle at which the ball can be hit, we can use the tangent function:
tan(θ) = vertical distance / horizontal distance
tan(θ) = 2.44 / 12.708
θ ≈ arctan(2.44 / 12.708)
θ ≈ 10.9 degrees
Therefore, the maximum angle (θ) at which the ball can be hit and land within the court is approximately 10.9 degrees.
To find the maximum angle at which the ball can be hit and land within the court, we can use the concept of projectile motion. Here's how you can calculate it:
Step 1: Determine the initial vertical velocity (Vy).
The initial vertical velocity (Vy) can be calculated using the formula:
Vy = √(2 * g * h)
where g is the acceleration due to gravity (approximately 9.8 m/s²) and h is the height from which the ball is spiked (2.44 meters in this case).
Vy = √(2 * 9.8 * 2.44)
Vy ≈ 6.25 m/s
Step 2: Calculate the maximum horizontal distance (R).
The maximum horizontal distance (R) can be calculated using the formula:
R = (Vy^2 / g)
R = (6.25^2 / 9.8)
R ≈ 3.99 meters
Step 3: Calculate the angle (θ).
The angle (θ) can be calculated using the formula:
θ = tan^(-1)(h / R)
θ = tan^(-1)(2.44 / 3.99)
θ ≈ 33.8 degrees
Therefore, to the nearest degree, the maximum angle (θ) at which the ball can be hit and land within the court is approximately 34 degrees.