Solving Volleyball Problem Homework w/ Max Speed & Angle

[agargento]

Homework Statement

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= 1.98 m directly above the back line, and the ball's initial velocity makes an angle θ = 39° with respect to the ground (see the figure). At what initial speed must the ball be hit so that it just barely makes it over the net? (Assume the volleyball is hit so that its path is parallel to the side-line as seen from an observer directly above the court, and that the volleyball is a point object.)

Homework Equations

V=V₀+at
r(t)=r₀+v₀t+½at²

The Attempt at a Solution

Well, there are six parts to this question. I did four of them.

(a. is in the question)
b. What is the maximum height above the court reached by the ball in this case?
c.At what initial speed must the ball be hit so that it lands directly on the opponent's back line?
d.What is the maximum height reached by the ball in this case?

Here are my answers:

a. 9.81 m/s
b. 3.92 m
c. 12.6 m/s
d. 5.19 m

I'm stuck at e. and f.

f. 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?

e. If you hit the ball at this maximum speed, at what angle should you strike it in order to make sure the ball lands in bounds?

--------------------

I don't even know where to start. First ,according to c. can the ball have an even higher speed? won't it get out of the field? If not, so how do I find the maxium speed it can have? I don't get how is this different then c.

e. Well... first we must find f ;)

Would love some help. Really stumped.

 

[jbriggs444]

f. 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 don't even know where to start. First ,according to c. can the ball have an even higher speed? won't it get out of the field? If not, so how do I find the maxium speed it can have? I don't get how is this different then c.

c assumes the launch angle is fixed and asks for the speed that results in the ball landing on the back line.

f take the launch angle as a variable (which will depend on launch speed) so that the ball just clears the top of the net. It then asks for the launch speed that results in the ball landing on the back line.

Beyond some maximum speed there is no launch angle that results in the ball both barely clearing the top of the net and also landing on the back line. [Unless, of course, the launch "contact point" is sufficiently high that there is a direct line of sight from contact point to backline that clears the net -- then you could serve as fast as you please. However, the given contact point in this case is not that high].

Edit:

 

[agargento]

c assumes the launch angle is fixed and asks for the speed that results in the ball landing on the back line.

f take the launch angle as a variable (which will depend on launch speed) so that the ball just clears the top of the net. It then asks for the launch speed that results in the ball landing on the back line.

Beyond some maximum speed there is no launch angle that results in the ball both barely clearing the top of the net and also landing on the back line. [Unless, of course, the launch "contact point" is sufficiently high that there is a direct line of sight from contact point to backline that clears the net -- then you could serve as fast as you please. However, the given contact point in this case is not that high].

Edit:

Ok, but I still don't understand how to start solving the problem.

 

[haruspex]

Ok, but I still don't understand how to start solving the problem.

You could start with some equations.
You have the usual equations for the horizontal and vertical directions. There is only one horizontal such, but a choice of five for vertical.
They have the common variable time, so you need to pick a vertical eqn involving time. You don't care about final velocity though.
But you don't care about time thereafter, so next step is to combine the two eqns to eliminate it.

 

[agargento]

You could start with some equations.
You have the usual equations for the horizontal and vertical directions. There is only one horizontal such, but a choice of five for vertical.
They have the common variable time, so you need to pick a vertical eqn involving time. You don't care about final velocity though.
But you don't care about time thereafter, so next step is to combine the two eqns to eliminate it.

For c. that is what I did. I can't figure out what's different this time... For c. I used the horizontal equation to express time, and then I put it in the equation for the vertical direction y(x)=x₀+v₀t-0.5at²...

 

[haruspex]

For c. that is what I did. I can't figure out what's different this time... For c. I used the horizontal equation to express time, and then I put it in the equation for the vertical direction y(x)=x₀+v₀t-0.5at²...

In c you knew the angle; now you do not. Please post the equation you get in terms of the unknown angle.

 

[agargento]

In c you knew the angle; now you do not. Please post the equation you get in terms of the unknown angle.

Well, in terms of the horizontal axis, we have 18=V₀cos(θ)t . In c I did t= 18/cos(θ) and substituted t in the equation y(x)=x₀+v₀t-0.5at² , and then I got only V₀ as a variable. But in c I knew the angle. Now I have 2 variables, V₀ and cos(θ)... Should I be doing something else or am I missing anything?

 

[haruspex]

Now I have 2 variables, V0 and cos(θ).

That's fine. You need the value of theta which maximises v₀. How do you find that?

 

[agargento]

That's fine. You need the value of theta which maximises v₀. How do you find that?

Make an equation that's equal to V₀ and then differentiate with respect to cos(θ)?

 

[haruspex]

Make an equation that's equal to V₀ and then differentiate with respect to cos(θ)?

It would be more usual to differentiate wrt θ, but that's the idea.

 

[agargento]

It would be more usual to differentiate wrt θ, but that's the idea.

What do you mean by that? What's the difference?

 

[haruspex]

What do you mean by that? What's the difference?

Either should work.

 

[agargento]

Either should work.

Well I started doing it and got some really insane equations... What exactly should I differentiate? I'm very confused.

 

[haruspex]

Well I started doing it and got some really insane equations... What exactly should I differentiate? I'm very confused.

I cannot tell where you are going wrong if you do not post your working.

 

[agargento]

I cannot tell where you are going wrong if you do not post your working.

Yeah you're right. I tried to separate V₀ from θ, but I think I missed the entire point completely.

 

[haruspex]

You need to resist the temptation to plug in numbers. Keep everything algebraic until the end. There are many advantages, not least that it makes it much easier to follow your working and spot mistakes.
Also, please do not post working as images. Take the trouble to type equations in. Guarantees legibility and makes it possible to quote individual lines in order to make comments. Images are for textbook extracts and diagrams.

Thinking this through again, you won't need any calculus. You can assume the max speed corresponds to just clearing the net and landing on the baseline. That gives you three equations with three unknowns: initial speed, launch angle and time to reach the net (i.e. half the total time).
Let the initial height be y₀, the net height y₁ and the court length be 2L (or whatever labels you prefer).
You can use unsubscripted v for initial speed since there are no other speeds in the equations.

 

[TomHart]

@agargento
One of the equations in your photo reads (assuming I am reading it correctly):
-1.98 = V_otanθ(18) - (4.9)(18)²/(V_ocosθ)²
That equation should be:
-1.98 = tanθ(18) - (4.9)(18)²/(V_ocosθ)²

In the first term on the right-hand side of the equation, the V_o should have canceled out with the V_o from the t = 18/V_ocosθ substitution - unless I'm mistaken.

And yes, as @haruspex mentioned, no calculus is necessary. You should have 3 equations: one for the x direction, and 2 for the y direction - one at the middle of the court (where the ball just clears the net) and one at the end of the court where the ball just drops in bounds. When you find the solution, please post the answer so I can check my work. :)

 

[PeroK]

Questions e) and f) seem to be much harder than the previous questions. Moreover, I can't see any short cuts based on the specific numbers, so a full general solution is required. No plug and chug here.

I was going to suggest we give the OP something to shoot at, in terms of the general solution for $v^2$ and $\tan(\theta)$.

 

[agargento]

Hey guys, sorry for disappearing, university's been rough. Anyway, me and a friend found a solution. I'll edit it in later. Thank you for all the help! Just wanted to let you know this is solved.

 

[PeroK]

I think this question would have been better if it asked you to show that, for e) and f):

$v^2 = \frac{g(L^2+(4d-3h)^2)}{4(2d-h)}$
and
$\tan \theta = \frac{4d-3h}{L}$

That would be something to shoot for.