A regulation volleyball court is L = 18.0 m long and a regulation volleyball net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.76 m directly above the back line, and the ball's initial velocity makes an angle q = 55° with respect to the ground.
From this information, I have been able to determine: 1) the initial speed the ball must be hit so that it just barely makes it over the net (9.96 m/s), 2) the maximum height above the court reached by the ball in that case (5.15 m), 3) the initial speed the ball must be hit so that it lands directly on the opponent's back line (13.3 m/s), and 4) the max. height reached by the ball in that case (7.77 m).
Now it says: "In volleyball, it is often advantageous to serve the ball as hard as possible. If you want the ball to land in the opponent's court, however, there is an upper limit on the initial ball speed for a given contact point. At this maximum speed, the ball just barely makes it over the net and then just barely lands in bounds on the back line of the opponent's court. For the contact point given in the previous problems, what is this maximum initial speed?"
I STILL am not getting this. Could someone maybe do the first few steps, and then I could see if I can get it after that??
Sure, I can help you with that. To find the maximum initial speed, we need to consider that the ball must barely make it over the net and then just barely land in bounds on the back line of the opponent's court.
Let's break down the problem into two parts:
1) The ball just barely makes it over the net.
2) The ball just barely lands in bounds on the back line.
Step 1: The ball just barely makes it over the net.
To find the maximum initial speed for the ball to clear the net, we can use the following approach:
We know that the height of the net (d) is 2.43 m and the ball's initial height (h) is 1.76 m. The angle of projection (q) is 55°.
To calculate the maximum initial speed (v) needed for the ball to clear the net, you can use the following equation:
v = sqrt((g * d) / (2 * sin^2(q) * cos^2(q))) + sqrt((g * h) / sin^2(q))
Where:
v = maximum initial speed
g = acceleration due to gravity (9.8 m/s²)
Let's calculate this:
v = sqrt((9.8 * 2.43) / (2 * sin^2(55°) * cos^2(55°))) + sqrt((9.8 * 1.76) / sin^2(55°))
Step 2: The ball just barely lands in bounds on the back line.
To find the maximum initial speed for the ball to land in bounds on the back line, we use the same approach as in Step 1. However, this time we take the initial height (h) as the maximum height reached by the ball in the previous case (7.77 m).
To calculate the maximum initial speed (v) needed for the ball to land in bounds on the back line, you can use the same equation as in Step 1.
v = sqrt((9.8 * 2.43) / (2 * sin^2(55°) * cos^2(55°))) + sqrt((9.8 * 7.77) / sin^2(55°))
Now, you can calculate both Step 1 and Step 2 to find the maximum initial speed for the ball to land in the opponent's court.
Sure! Let's break down the problem step by step.
First, let's determine the requirements for the ball to just barely make it over the net and then just barely land in bounds on the back line of the opponent's court.
To do this, we need to consider the initial height at which the ball is struck, the angle at which it is struck, and the maximum height reached by the ball.
From the given information, we know that the ball is struck at a height h = 1.76 m directly above the back line, and the angle of projection is q = 55° with respect to the ground. We also know the maximum height reached by the ball is 7.77 m.
To calculate the maximum initial speed needed, we can use the equations of projectile motion. The key equation we'll use is:
v^2 = u^2 + 2as
where:
v = final velocity of the ball (which is zero when it reaches the maximum height and when it lands),
u = initial velocity of the ball,
a = acceleration due to gravity (which is approximately -9.8 m/s^2),
s = displacement (which is the difference in height between the initial and final positions of the ball).
Let's analyze the different stages of the ball's motion:
Stage 1: As the ball moves upward, its vertical displacement (s1) is equal to the maximum height reached by the ball (7.77 m). The acceleration due to gravity is acting against its motion, so the displacement is positive while the acceleration is negative.
Stage 2: At the topmost point of its trajectory, the ball momentarily comes to rest, so its final velocity (v2) is zero. The displacement (s2) in this stage is equal to the vertical height at which the ball was struck (1.76 m). The acceleration is still negative.
Stage 3: The ball falls downward towards the opponent's court. The displacement (s3) in this stage is equal to the vertical height at which the ball was struck (1.76 m) minus the maximum height reached by the ball (7.77 m). The acceleration due to gravity is acting in the same direction as the motion, so the displacement is negative while the acceleration is positive.
Using these values, we can now calculate the maximum initial speed needed for the ball to just barely make it over the net and then just barely land in bounds on the back line of the opponent's court.
Let's denote the maximum initial speed as u_max.
To find u_max, we need to find the initial velocity that allows the ball to complete all three stages. This can be done by combining the equations for the three stages and solving for u.
I'll provide the steps for solving this equation. Please let me know when you feel comfortable continuing from that point.