What Initial Speed Is Required for a Volleyball to Just Clear the Net?

In summary, to hit the ball just barely over the net, the ball must be hit with an initial speed of 9.97 m/s at an angle of 49 degrees with respect to the ground. This is calculated by finding the vertical and horizontal components of velocity, using the x and y motion equations, and eliminating one variable to solve for the other.
  • #1
trumnation
3
0

Homework Statement



A regulation volleyball court is L = 18.0 m long and the net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.61 m directly above the back line, and the ball's initial velocity makes an angle θ = 49° with respect to the ground. At what initial speed must the ball be hit so that it just barely makes it over the net? (Assume the volleyball is a point object and is hit so that its path is parallel to the sideline, as seen from directly above the court.)

The Attempt at a Solution



I am really lost at how to figure out this one without time. I tried using the following equation since you are given the height it has to reach, and the acceleration (-9.8) and I think the final velocity would be zero:
[tex]
v^2 = v_0^2 + 2 a \Delta x
[tex]

but it doesn`t seem to work. I am thinking I have to calculate the time somehow, but I really can`t figure out how to go about doing that. Can anyone help?
 
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  • #2
trumnation said:

Homework Statement



A regulation volleyball court is L = 18.0 m long and the net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.61 m directly above the back line, and the ball's initial velocity makes an angle θ = 49° with respect to the ground. At what initial speed must the ball be hit so that it just barely makes it over the net? (Assume the volleyball is a point object and is hit so that its path is parallel to the sideline, as seen from directly above the court.)

The Attempt at a Solution



I am really lost at how to figure out this one without time. I tried using the following equation since you are given the height it has to reach, and the acceleration (-9.8) and I think the final velocity would be zero:

[tex] v^2 = v_0^2 + 2 a \Delta x [/tex]

but it doesn`t seem to work. I am thinking I have to calculate the time somehow, but I really can`t figure out how to go about doing that. Can anyone help?

It's not quite so simple.

First you need to determine the Vertical and Horizontal components of velocity based on the 49 degrees.

Then you can develop the x and y motion equations using the 18 m as the range and the height of the net as the height at the instant of clearance.

The equations

[tex] Y_{net} = Y_{serve} +V_y*t -1/2*g*t^2[/tex]

[tex] X_{to.net} = V_x*t [/tex]
 
  • #3
Ok, how though, do I find the time? Or should I solve for t in the x equation and sub it in?
 
  • #4
trumnation said:
Ok, how though, do I find the time? Or should I solve for t in the x equation and sub it in?

You should have equations with only initial velocity and time unknown.

Eliminate 1 and solve for the other.
 
  • #5
Ok so I tried that, and I think I have an error somewhere because it doesn't work. Here is what I did:

t = 9/(Vsin49)

2.43 = 1.61 + Vcos49t - 0.5*9.8t^2
0.82 = Vcos49 * 9/Vsin49 - 4.9 * (9/Vsin49)^2

V = 7.9 m/s

I can't seem to find what I did wrong though..
 
  • #6
trumnation said:
Ok so I tried that, and I think I have an error somewhere because it doesn't work. Here is what I did:

t = 9/(Vsin49)

2.43 = 1.61 + Vcos49t - 0.5*9.8t^2
0.82 = Vcos49 * 9/Vsin49 - 4.9 * (9/Vsin49)^2

V = 7.9 m/s

I can't seem to find what I did wrong though..

I calculate your equation differently.

Using Sin49 = .755 and Cos49=.656
I get: t=11.92/v

Substituting:
.82 = .656(11.92) - 4.9(11.92/V)2
.82 = 7.82 - 4.9(11.92/V)2
7 = 4.9(11.92/V)2
V2 = 4.9(11.92)2/7 = 99.46
V = 9.97 m/s
 

FAQ: What Initial Speed Is Required for a Volleyball to Just Clear the Net?

1. What is projectile motion?

Projectile motion is the motion of an object that is moving through the air or space under the influence of gravity. In the case of volleyball, the ball is the projectile and it follows a curved path as it travels through the air.

2. How does projectile motion apply to volleyball?

In volleyball, projectile motion is seen when a player serves, spikes, or passes the ball. The ball is launched into the air at an angle and follows a parabolic path, determined by its initial velocity and the force of gravity.

3. What factors affect the trajectory of a volleyball in projectile motion?

The trajectory of a volleyball is affected by the initial speed and angle at which it is launched, as well as the force of gravity and air resistance. Other factors such as wind and spin on the ball can also impact its trajectory.

4. How can we calculate the distance and height of a volleyball in projectile motion?

To calculate the distance and height of a volleyball in projectile motion, we can use the equations of motion, which take into account the initial velocity, angle of launch, and time of flight. By plugging in these values, we can determine the horizontal distance and maximum height reached by the ball.

5. How does air resistance affect the motion of a volleyball in projectile motion?

Air resistance, also known as drag, can affect the trajectory of a volleyball in projectile motion by slowing down its horizontal and vertical velocities. This means that the ball will not travel as far and will not reach as high of a maximum height as it would without air resistance. The impact of air resistance is greater for slower moving objects, so it is more noticeable in volleyball serves and passes rather than spikes.

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