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?"
What equation can I use to solve this? I've tried a couple of different ways before, and nothing seems to be working.
Use the vertical displacement equation to find the time of flight in terms of initial velocity. Then use the horizontal displacement equation with that expression for time to find the initial velocity.
To solve this problem, you can use the range equation for projectile motion. The range equation relates the initial velocity, launch angle, and time of flight to the horizontal distance traveled by the projectile. In this case, you want to find the maximum initial speed such that the ball just barely makes it over the net and lands on the opponent's back line.
The range equation is given by:
R = (v^2 * sin(2θ)) / g
Where:
- R is the horizontal distance traveled by the projectile
- v is the initial velocity of the projectile
- θ is the launch angle of the projectile
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
In this case, the horizontal distance R you are interested in is the length of the volleyball court, which is 18.0 m. The launch angle θ is given as 55°.
Now, let's rearrange the equation to solve for the initial velocity v:
v^2 = (R * g) / sin(2θ)
v = √((R * g) / sin(2θ))
Substituting the known values, we get:
v = √((18.0 * 9.8) / sin(2 * 55°))
v ≈ 20.8 m/s
Therefore, the maximum initial speed the ball can have to just barely clear the net and land in bounds on the opponent's back line is approximately 20.8 m/s.