What Calculations Determine the Volleyball's Trajectory in Projectile Motion?

• lolseeker2
In summary: If AC=CB and the ball is at its highest point at C, isn't at point B going to be also at height h above the ground? (air resistance...neglecting it)No, there would be a difference in height between point B and point C.
lolseeker2
Member warned that the homework template must be used
The volleyball player serves the ball from pt.A w/ an initial Vo at angle theta to the horizontal. If the ball just cleared the net at pt. C and landed on the base at line B.

Determine the value of theta.

Determine the value of Vo.

Determine the time reach to the highest point at the reference from the horizontal surface.

The h from which the ball is thrown is 0.9 m.

https://imgur.com/a/qDkzsbO

I tried solving Vox using Vox = 0.9sin90 by assuming the other angle from Vox to h to be 90 degrees.

I don't know how to get Voy to solve for the theta.

We can assume that the distances AB,AC,CB as well as the height of point C from the ground, are all known to us?

Delta² said:
We can assume that the distances AB,AC,CB as well as the height of point C from the ground, are all known to us?
The only variable that has a value is the height relative to the ground from which the ball is thrown by the player which is 0.9m

lolseeker2 said:
Vox = 0.9sin90
The 0.9 (m) is a distance, so that is not going to produce a velocity. I have no idea what
lolseeker2 said:
the other angle from Vox to h to be 90 degrees.
means.
As Δ2 points out, you need to know the dimensions of a volleyball court. Look those up.

You should not have deleted the template. What relevant standard equations do you know for motion under constant acceleration?

haruspex said:
The 0.9 (m) is a distance, so that is not going to produce a velocity. I have no idea what

means.
As Δ2 points out, you need to know the dimensions of a volleyball court. Look those up.

You should not have deleted the template. What relevant standard equations do you know for motion under constant acceleration?

Ah yes...my mistake...should have considered that. What equations should i use for solving theta?

Something else, when you say "it just clears the net at point C" , do you mean that the ball is at its highest point at point C?

Delta² said:
Something else, when you say "it just clears the net at point C" , do you mean that the ball is at its highest point at point C?
yes

Something doesn't look quite right... For a volleyball court it is AC=CB right?

or point A is a bit out than the baseline at that side, while point B is at the baseline?

Delta² said:
Something doesn't look quite right... For a volleyball court it is AC=CB right?
yes...i just did a diagram of it...I also assumed the height for the net is for men's.

If the ball is at its highest point at point C, and AC=CB then at point B it will be at height h from the ground (its from the symmetry of the situation, if it spends horizontal distance AC to reach the highest point, then will also spend another AC=CB distance to reach at height h which is the height it is initially launched).
It can't land exactly there, at point B, regardless of what we have for ##v_0## and ##\theta##

Delta² said:
Something doesn't look quite right... For a volleyball court it is AC=CB right?

or point A is a bit out than the baseline at that side, while point B is at the baseline?

Point B is at the baseline. Pt. A itself is where the player throws the ball with a height of 0.9 m from the ground.

is Point A exactly above the baseline at that side and at height 0.9m?

The ball was thrown from pt. A via player's hand at a height of 0.9m from the ground. The ball reach max height at pt. C when it cleared the net and then it dropped exactly at pt. B baseline. That's how I pictured out problem.

Delta² said:
is Point A exactly above the baseline at that side and at height 0.9m?
yes

lolseeker2 said:
The ball was thrown from pt. A via player's hand at a height of 0.9m from the ground. The ball reach max height at pt. C when it cleared the net and then it dropped exactly at pt. B baseline. That's how I pictured out problem.
This scenario isn't possible unless we account for air resistance. Are we to neglect air resistance or not?

Delta² said:
This scenario isn't possible unless we account for air resistance. Are we to neglect air resistance or not?
yes

if we neglect air resistance, you scenario just cannot happen.

Well the teacher who gave the problem stated there is no air resistance involved.

@haruspex what do you think, if AC=CB and the ball is at its highest point at C, isn't at point B going to be also at height h above the ground? (air resistance neglected)

I know of a way to solve this. We have three points on a parabola: (0, 0.9), (9, 2.43), (18, 0). One can find the equation of the parabola which can lead to the answer. Point C will not be the highest point, it'll peak just before point C.

So if you want to follow this method, start by finding the coefficients of y = ax^2 + bx + c.

Delta2
verty said:
I know of a way to solve this. We have three points on a parabola: (0, 0.9), (9, 2.43), (18, 0). One can find the equation of the parabola which can lead to the answer. Point C will not be the highest point, it'll peak just before point C.

So if you want to follow this method, start by finding the coefficients of y = ax^2 + bx + c.
If the ball peaks before point C then I agree it is solvable. Otherwise we have to take into account air resistance (drag).

Delta² said:
@haruspex what do you think, if AC=CB and the ball is at its highest point at C, isn't at point B going to be also at height h above the ground? (air resistance neglected)
Should not assume it is at the highest point at the net. That is not implied by "just clears the net".
We have three (x,y) points that we know it passes through. That gives us four equations. Correspondingly, there are four unknowns: the initial velocity components, the time to reach the net, and the time to reach the baseline.

Ok, I can see Haruspex's method is the one to use (for sure).

verty said:
Ok, I can see Haruspex's method is the one to use (for sure).

(Not sure) , your method involves at start solving a system of 3 linear equations, which i guess won't be that hard, then we can find theta by ##\tan\theta={\frac{dy}{dx}}_{x=0}##. And then we can find the ##x_0## for which we have the peak by solving ##\frac{dy}{dx}=0## (also a linear equation) and then we can find the ##y_0## (highest height) and then get ##v_0## from ##v_0\sin\theta=\sqrt{2gy_0}## (also linear equation) and the time of ascent ##t_0##, by solving another linear equation ##x_0=v_0t_0\cos\theta##.

Haruspex method involves solving at least 2 quadratic equations if I am not mistaken.

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Delta² said:
Haruspex method involves solving at least 2 quadratic equations if I am not mistaken.
It is not necessary to solve any quadratic equations, just two (or three, depending on how you choose to write them) simultaneous equations which are effectively linear.
The way I expressed it in post #23 was not intended to represent the best approach, merely that there is the right number of equations for the unknowns without assuming the highest point is at the net.

I still think your methods involves solving a system of two linear and two quadratic equations... What do you mean by the term simultaneous equations?

Delta² said:
I still think your methods involves solving a system of two linear and two quadratic equations... What do you mean by the term simultaneous equations?
In its simplest form, two linear equations with two unknowns. In this case, it is quadratic in one unknown, but to find the angle you only need to solve for the other one.

haruspex said:
In its simplest form, two linear equations with two unknowns. In this case, it is quadratic in one unknown, but to find the angle you only need to solve for the other one.
Not sure what you mean here, but I agree if you correctly treat the system , you can solve one quadratic equation and some linear equations and get the angle and the velocities.

Delta² said:
(Not sure) , your method involves at start solving a system of 3 linear equations, which i guess won't be that hard, then we can find theta by ##\tan\theta={\frac{dy}{dx}}_{x=0}##. And then we can find the ##x_0## for which we have the peak by solving ##\frac{dy}{dx}=0## (also a linear equation) and then we can find the ##y_0## (highest height) and then get ##v_0## from ##v_0\sin\theta=\sqrt{2gy_0}## (also linear equation) and the time of ascent ##t_0##, by solving another linear equation ##x_0=v_0t_0\cos\theta##.

Haruspex method involves solving at least 2 quadratic equations if I am not mistaken.

I believe Haruspex's method does not require one to differentiate at all which I think is much nicer.

verty said:
I believe Haruspex's method does not require one to differentiate at all which I think is much nicer.
That's very true, I still like more your method for that exact reason, that it involves derivatives!

Delta² said:
Not sure what you mean here, but I agree if you correctly treat the system , you can solve one quadratic equation and some linear equations and get the angle and the velocities.
You get two linear equations in which the unknowns are vy/vx and 1/vx2, vx and vy being the initial velocity components.

Delta2
verty said:
Ok, I can see Haruspex's method is the one to use (for sure).
It is no different from your own method in post #21. For some reason I had not seen that when I wrote post #23. There can be quite a delay.
I get two equations instead of three merely by taking coordinates relative to the launch point.

haruspex said:
It is no different from your own method in post #21. For some reason I had not seen that when I wrote post #23. There can be quite a delay.
I get two equations instead of three merely by taking coordinates relative to the launch point.

I meant working with the usual time formula. Solving those quadratics was not so difficult in this case.

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1. How do I determine the initial velocity and angle for a projectile motion problem?

The initial velocity and angle can be determined by using the given information in the problem. The initial velocity can be found using the equation v0 = vxcosθ, where v0 is the initial velocity, vx is the velocity in the x-direction, and θ is the angle of launch. The angle can be found using the equation tanθ = vy/vx, where vy is the velocity in the y-direction.

2. How do I find the maximum height and range of a projectile?

The maximum height and range can be found by using the equations hmax = (v02sin2θ)/2g and R = (v02sin2θ)/g, where hmax is the maximum height, R is the range, v0 is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity (9.8 m/s2).

3. How do I account for air resistance in a projectile motion problem?

To account for air resistance, you can use the equation FD = 0.5ρAv2, where FD is the drag force, ρ is the density of the fluid, A is the cross-sectional area of the object, and v is the velocity of the object. This equation can be incorporated into the equations of motion to calculate the trajectory of the projectile.

4. How do I calculate the time of flight for a projectile?

The time of flight can be calculated using the equation t = 2v0sinθ/g, where t is the time of flight, v0 is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity. This equation assumes that the object is launched from and lands at the same height.

5. How do I solve for the velocity and displacement at a specific time in a projectile motion problem?

To solve for the velocity and displacement at a specific time, you can use the equations v = v0 + at and x = x0 + v0t + 0.5at2, where v is the velocity, v0 is the initial velocity, a is the acceleration, x is the displacement, x0 is the initial displacement, and t is the time. Plug in the given time value to solve for the velocity and displacement at that time.

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