Homework Help: Volleyball problem

1. Dec 2, 2016

agargento

• Member advised to arrange the post content into the appropriate template sections
1. The problem statement, all variables and given/known data

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.)

2. Relevant equations

V=V0+at
r(t)=r0+v0t+½at2

3. 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?

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.

2. Dec 2, 2016

jbriggs444

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:

3. Dec 3, 2016

agargento

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

4. Dec 3, 2016

haruspex

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.

5. Dec 3, 2016

agargento

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...

6. Dec 3, 2016

haruspex

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

7. Dec 3, 2016

agargento

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?

8. Dec 4, 2016

haruspex

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

9. Dec 4, 2016

agargento

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

10. Dec 4, 2016

haruspex

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

11. Dec 4, 2016

agargento

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

12. Dec 4, 2016

haruspex

Either should work.

13. Dec 4, 2016

agargento

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

14. Dec 4, 2016

haruspex

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

15. Dec 4, 2016

agargento

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

16. Dec 4, 2016

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.

17. Dec 4, 2016

TomHart

@agargento
-1.98 = Votanθ(18) - (4.9)(18)2/(Vocosθ)2
That equation should be:
-1.98 = tanθ(18) - (4.9)(18)2/(Vocosθ)2

In the first term on the right-hand side of the equation, the Vo should have cancelled out with the Vo from the t = 18/Vocosθ 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. :)

18. Dec 5, 2016

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 $\tan(\theta)$.

19. Dec 5, 2016

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.

20. Dec 5, 2016

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.