With what minimum speed does the ball need to be hit with to go over the net

[~christina~]

[SOLVED] With what minimum speed does the ball need to be hit with to go over the net

Homework Statement

During the semi-finals of the U. S. Open 2007 Grand Slam Tennis, Venus Williams was in the second set of her match against Justine Henin when her serve was clocked (by a radar gun) at 77 miles per hour.
The ball left her racquet horizontally at a height of 2.37 m above the surface of the tennis court. Venus was standing at the service line 12.0 m from the net that has a height of 0.9 m.

If the ball clears the net and lands within 7.0 m of the net on the other side, the serve is considered “good" according to the regulations of the game because the ball lands inside the service box on the other side of the net. (See the sketch of a tennis court below.) If the ball clears the net but lands beyond the service box on the other side, this serve is considered a fault. The player receiving the service ball cannot hit the ball before it strikes the court inside the service box.

(a) With what minimum speed (in SI units) is Venus required to strike the ball for it to just clear the net? State assumption(s) and support your response quantitatively.

(b) Was the serve clocked at 77 miles per hour "good" or was it a fault? If this ball goes over the net, how far will it land from the backline of the service box on Justine’s side of the net. Support your response quantitatively.

(c) Sketch graphs of the ball’s displacement, velocity, and acceleration as functions of time (six graphs.)

(d) Williams’ slowest serve was at 71 miles per hour. She used the same technique described above but she obviously did not strike this ball as hard. After the match, a local sportscaster said "By serving a slower ball that cleared the net, Williams gave Henin more time to react to her serve because the ball was in the air a little longer than when Williams served her fast ball". Does the sportscaster speak with authority on this subject? Use physics principles to support your response.

the picture looks like this:
person at one end
__________12ft_______________|[_________7ft________]______

Homework Equations

I don't really know...is it projectile motion??

The Attempt at a Solution

well I know that her hit is

v= 77m/hr wouldn't I have to convert that or something since velocity is usually m/s?

dist from net = 12m
net height= 0.9m high

y component for ball is = 2.37m

I'm not sure where to start this problem...

Can you help me out?


~Thanks~

 

[asd1249jf]

Yes, this would be considered as a projectile motion problem, with an initial given height.

Let's take this step by step.

Recall the following equations for such type of problems

[tex]v_x = v_0\cos(\theta)[/tex]
[tex]v_y = v_0\sin(\theta) - gt[/tex]
[tex]s_x = s_0x + v_xt[/tex]
[tex]s_y = s_0y + v_yt - 0.5gt^2[/tex]

First off, you know the ball was slammed horizontally so you would have [tex]\theta = 0[/tex] degrees.

For part a, you know the height of the net and the horizontal distance of the net with respect to Venus Williams. Plug in the height of the net into the [tex]s_y[/tex] equation to find the time it takes for the ball to reach the net. Specifically speaking, you are actually finding the time which your ball takes to drop from your initial height (In this case, the height of the ball above the ground as it was struck) to the height of net. Leave [tex]v_0[/tex] as a variable in your calculation. (your [tex]s_0y[/tex] is the initial height and from the problem description, you know there is no initial y-component of velocity, so you only have gravity component and initial height left in your [tex]s_y[/tex] equation). Now, plug in the time which the ball took to travel to your desired point and your horizontal distance into the [tex]s_x[/tex] equation. You will now be able to solve the initial velocity, asked in part a.

Come back when you have the answer ready.

 

[~christina~]

Okay well let's see if I'm alright in my thinking and work and If I understood you correctly.

x from net = 12ft= 3.657m
service line x= 7ft from net
so ball can't go over 19ft or 5.7912m

ball angle = 0 ( I couldn't find the theta symbol)

vi is ?

you say to plug into the equation
so doing that:

sy= soy + vy + 0.5gt^2

0.9m= 2.37m + voy - 4.9 t^2

you said to leave voy as a variable...does that mean it's just ignored? I know I don't have that but I can just leave it out?? Is it 0?
well taking that out I get:

-1.47= -1.49t^2
t^2= 0.3
t= 0.548s

taking that and and plugging the t = 0.548s ( if I'm correct..the time it takes for the ball to reach the net) ( but if it needs to clear the net wouldn't it have to be just higher than the net??)

now you said to plug into the sx= sox + vx*t equation with this time to see the velocity it takes to hit the ball that far in the direction of the x axis.

so I converted the values since they have to be in SI units according to the question..

x values
x from net = 12ft= 3.657m
service line x= 7ft from net
t=
so using those values in that equation.

sx=sox + vx*t

3.657m= 0 + vx * 0.548s

vx= 6.67m/s

This would be the time it takes to reach the net and just clear it right?
________________________________________________________________
Q a.):State assumption(s) and support your response quantitatively.

~what would that mean to state the assumption? do I guess at first and then show what I found ? and how do I support it quantitatively? I assume that would mean to show my work and why I did what I did. Am I correct?

_________________________________________________________________

Well You said to come back and post what I did after I did it so now how do I do b.) assuming part a.) is okay for what I did?

b.) (b) Was the serve clocked at 77 miles per hour "good" or was it a fault? If this ball goes over the net, how far will it land from the backline of the service box on Justine’s side of the net. Support your response quantitatively.


~Thanks~

 

[asd1249jf]

Okay well let's see if I'm alright in my thinking and work and If I understood you correctly.

x from net = 12ft= 3.657m
service line x= 7ft from net
so ball can't go over 19ft or 5.7912m

ball angle = 0 ( I couldn't find the theta symbol)

vi is ?

Well first of all, your problem description states that [tex]s_ox[/tex] is 12m, not 12 ft.

you say to plug into the equation
so doing that:

sy= soy + vy + 0.5gt^2

0.9m= 2.37m + voy - 4.9 t^2

On your equation

0.9m= 2.37m + voy - 4.9 t^2

Remember that the ball was slammed horizontally. Meaning that there is no velocity in y direction, so that's actually 0 (And it seems like that's what you did). You can solve for time, which is the time for the ball it takes to drop from the initial height to the height of the net.

-1.47= -1.49t^2
t^2= 0.3
t= 0.548s

Looks like you got the time right.

taking that and and plugging the t = 0.548s ( if I'm correct..the time it takes for the ball to reach the net) ( but if it needs to clear the net wouldn't it have to be just higher than the net??)

No. I'll explain why later. Keep on reading.

now you said to plug into the sx= sox + vx*t equation with this time to see the velocity it takes to hit the ball that far in the direction of the x axis.

so I converted the values since they have to be in SI units according to the question..

x values
x from net = 12ft= 3.657m
service line x= 7ft from net
t=
so using those values in that equation.

sx=sox + vx*t

3.657m= 0 + vx * 0.548s

vx= 6.67m/s

This would be the time it takes to reach the net and just clear it right?

Again, from your problem description, the horizontal distance is 12m. You might want to double check this. But as far as your method of calculation goes, it is correct.

Q a.):State assumption(s) and support your response quantitatively.

~what would that mean to state the assumption? do I guess at first and then show what I found ? and how do I support it quantitatively? I assume that would mean to show my work and why I did what I did. Am I correct?

Stating assumption means that you have calculated the values ignoring certain criterias that may happen in the real world. For instance, the air resistance - We are assuming that there is none. Also, as mentioned earlier, you asked about the ball needing to be slightly higher than the net. In the real world, that is correct however, we do not have information about the size of the ball. For that reason, we are assuming the ball is infinitesmally small and it clears the net just above it.

Quantitatively supporting your answer means just doing the calculation with numbers, which is exactly what you did above.

Well You said to come back and post what I did after I did it so now how do I do b.) assuming part a.) is okay for what I did?

b.) (b) Was the serve clocked at 77 miles per hour "good" or was it a fault? If this ball goes over the net, how far will it land from the backline of the service box on Justine’s side of the net. Support your response quantitatively.

First, convert your minimum velocity you've found earlier to miles/per hour. Is it less than the serve clocked at 77 miles per hour? If yes, then you know that the ball didn't make it over the net since Venus struck the ball at a slower speed than what is required.

If not, then you need to determine where, on the otherside the ball lands in order to find out if the serve is considered "Good" or not.

Again, referring back to the following equations

[tex]s_x = s_x0 + v_xt[/tex]
[tex]s_y = s_y0 + v_yt - 0.5gt^2[/tex]

Notice to find the horizontal distance, you first need to find the time which it takes for the ball to land on the ground. Again, your y component of velocity is 0.

Apparently, your y-distance has to equal to 0 when the ball hits the ground. Plug in 0 for the y distance, and you've given the initial height so you can solve for time now.

Now, you can use the time you just found and plug in into the horizontal distance equation (With initial horizontal distance = 0) to find how far the ball has traveled with respect to venus.

To determine if the serve was good or bad, you need to see if the ball landed within 7 m (or feet, please check) in the opponent's side. The horizontal distance you've calculated above is with respect to venus, so to find the distance from the net to the backline of the opponent's side, you have to subtract the calculation by the distance between venus and the net, which is 12 (ft or m). (If you get a negative value, then you have done something wrong). If the result is greater than 7m, the serve is a fault. If not, then it is good.

Lastly, you have been asked to find the distance from the backline of the Justine's side of the net. To find this, you can simply subtract the total distance, which is from venus to the backline of justine's side by the distance of the ball with respect to Venus.

 

[~christina~]

First, convert your minimum velocity you've found earlier to miles/per hour. Is it less than the serve clocked at 77 miles per hour? If yes, then you know that the ball didn't make it over the net since Venus struck the ball at a slower speed than what is required.

If not, then you need to determine where, on the otherside the ball lands in order to find out if the serve is considered "Good" or not.

well I corrected the distance to m and I found that the v= 21.90m/s and in mi/hr
and got 48.99mi/hr

you say that if it is less than the serve clocked at 77mi/hr then it didn't make it over the net but Isn't I don't get that since I found that the speed needed for the ball to go over the net is less than what the ball was hit with and since it was hit with more speed than is needed for the ball to cross the net then it should go over the net

since my reasoning is that 77mi/hr= 34.42m/s (the serve)
while the minimum needed to clear the net was 48.99mi/hr = 21.90m/s and thus her hit had a higher velocity than what was needed by 12.52m/s
__________
Assuming my thinking is correct I now find the distance it falls from the net at the 77mi/hr serve

Sx= Soy + vxt
Sy= Soy + vyt + 0.5gt^2

vx= 34.42m/s ==> velocity of serve
g= -9.8
vyt= 0
Soy= 2.37m
Sy= 0

0= 2.37m + 0*t + 0.5(-9.8)t^2
-2.37m = -4.9t^2
t= 0.695s ===> to reach ground

_________________________
now to find the distance the ball has traveled with respect to Venus

Sx= Soy + Vxt
Sx= 0 + (34.42m/s)(0.695s)
Sx= 23.92m

You said to subtract to find the distance it travels after crossing the net is 12m
so
distance traveled after crossing net : 23.92m- 12m= 11.92m

would definitely go over the 7m mark for the backline.

Is this fine?
So it would be a fault I guess
_________________________

For the distance from the backline the ball lands after I did this...

7m => backline distance
12m => distance from Venus to the net
so total = 19m

23.56-19m= 4.92from the backline.
________________________
(c) Sketch graphs of the ball’s displacement, velocity, and acceleration as functions of time (six graphs.)

which equation would I use?? wouldn't the graph's acceleration as a function of time be the Jerk? and is the reason it is 6 graphs is b/c each represents 1 axis like the x-axis motion velocity vs time graph. And what numbers would I use for the graph?? I know that for Sy= Soy + Vyt + 0.5gt^2 Soy wouldn't there only be one value that can change since it's a graph ? so if I was doing the velocity vs time graph would it be that I would use Sx = Soy + Vxt where for the x-axis (horizontal motion) and initial height would be 2.37m and but I'm confused after that...

(d) Williams’ slowest serve was at 71 miles per hour. She used the same technique described above but she obviously did not strike this ball as hard. After the match, a local sportscaster said "By serving a slower ball that cleared the net, Williams gave Henin more time to react to her serve because the ball was in the air a little longer than when Williams served her fast ball". Does the sportscaster speak with authority on this subject? Use physics principles to support your response

71mi/hr = 31.74m/s well I guess I would just find the time it takes to land to see if it really is in the air longer.
Assuming it is correct (my thought)

vx= 31.74m/s
Vyt= 0
Soy= 2.37m
Sy= 0
g= -9.8m/s^2

Sy= Soy + Vyt + 0.5gt^2
0= =2.37m + 0*t + 0.5(-9.8)t^2
-2.37= -4.9t^2
t= 0.695
0= 2.37 + 0*t + 0.5 (-9.8)t^2
t= 0.695s

~Then again after thinking about this it doesn't matter how fast you serve the ball b/c that only affects the motion in the x direction and the time the ball stays in the air is the same so the commentator is incorrect in physics principles.


~The most important thing that I'm asking is the graphing equations that I use to graph the velocity vs time, acceleration vs time, and displacement vs time~

I really need to finish this question by today thank you

 

[asd1249jf]

well I corrected the distance to m and I found that the v= 21.90m/s and in mi/hr
and got 48.99mi/hr

you say that if it is less than the serve clocked at 77mi/hr then it didn't make it over the net but Isn't I don't get that since I found that the speed needed for the ball to go over the net is less than what the ball was hit with and since it was hit with more speed than is needed for the ball to cross the net then it should go over the net

since my reasoning is that 77mi/hr= 34.42m/s (the serve)
while the minimum needed to clear the net was 48.99mi/hr = 21.90m/s and thus her hit had a higher velocity than what was needed by 12.52m/s

OOPS, sorry, I wasn't thinking clearly. You are totally right - Since she hit the ball at a faster rate than what is required for the ball to make it over the net, the ball at least clears the net.

Assuming my thinking is correct I now find the distance it falls from the net at the 77mi/hr serve

Sx= Soy + vxt
Sy= Soy + vyt + 0.5gt^2

vx= 34.42m/s ==> velocity of serve
g= -9.8
vyt= 0
Soy= 2.37m
Sy= 0

0= 2.37m + 0*t + 0.5(-9.8)t^2
-2.37m = -4.9t^2
t= 0.695s ===> to reach ground

_________________________
now to find the distance the ball has traveled with respect to Venus

Sx= Soy + Vxt
Sx= 0 + (34.42m/s)(0.695s)
Sx= 23.92m

You said to subtract to find the distance it travels after crossing the net is 12m
so
distance traveled after crossing net : 23.92m- 12m= 11.92m

would definitely go over the 7m mark for the backline.

Is this fine?
So it would be a fault I guess

As long as you plugged in the numbers into the calculator correctly, your method of calculation is correct.

For the distance from the backline the ball lands after I did this...

7m => backline distance
12m => distance from Venus to the net
so total = 19m

23.56-19m= 4.92from the backline.

Looks correct.

________________________
(c) Sketch graphs of the ball’s displacement, velocity, and acceleration as functions of time (six graphs.)

which equation would I use?? wouldn't the graph's acceleration as a function of time be the Jerk? and is the reason it is 6 graphs is b/c each represents 1 axis like the x-axis motion velocity vs time graph. And what numbers would I use for the graph?? I know that for Sy= Soy + Vyt + 0.5gt^2 Soy wouldn't there only be one value that can change since it's a graph ? so if I was doing the velocity vs time graph would it be that I would use Sx = Soy + Vxt where for the x-axis (horizontal motion) and initial height would be 2.37m and but I'm confused after that...

Jerk is given as change of acceleration over time. If you know calculus, this is the derivative of acceleration. So to respond to your question - no, graphing acceeration as a function of time would not be considered as jerk. BTW, careful with your [tex]S_x[/tex] equation, that doesn't involve initial height. As for graphing the displacement, you need to graph [tex]S_x[/tex] and [tex]S_y[/tex] against time. As for [tex]S_y[/tex] graph, plot it until [tex]S_y[/tex] reaches 0 and as for [tex]S_x[/tex], plot it until the time where [tex]S_y[/tex] reaches 0. You know how to graph certain functions such as [tex]4x^2 + 3[/tex], right? It's the same idea - All you have to do to graph this is to treat [tex]S_x[/tex] and [tex]S_y[/tex] as Y, and your time as X. Same goes for the velocity equations given above, of x component and y component. For example - If you wanted to graph [tex]S_x[/tex] vs. Time, you take your equation, [tex]S_x = 34.42t[/tex] (Here, I am assuming [tex]v_x[/tex] to be 34.42m/s, since the problem didn't specify which particular serve to graph) plot [tex]s_x[/tex] on the y-axis and t on your x axis. You would get a diagonal line, since this is a linear equation.

Acceleration is the easiest to graph of all - As for the y component of acceleration, you know there has to be only ONE component of acceleration acting on the ball - the gravity, and you know the acceleration due to gravity acts constantly on the ball. As for the x component of acceleration, review your [tex]v_x[/tex] equation. Notice that the x component of velocity is constant throughout the motion - Then ask yourself this question : What is the acceleration of an object that has a constant velocity?

(d) Williams’ slowest serve was at 71 miles per hour. She used the same technique described above but she obviously did not strike this ball as hard. After the match, a local sportscaster said "By serving a slower ball that cleared the net, Williams gave Henin more time to react to her serve because the ball was in the air a little longer than when Williams served her fast ball". Does the sportscaster speak with authority on this subject? Use physics principles to support your response

71mi/hr = 31.74m/s well I guess I would just find the time it takes to land to see if it really is in the air longer.
Assuming it is correct (my thought)

vx= 31.74m/s
Vyt= 0
Soy= 2.37m
Sy= 0
g= -9.8m/s^2

Sy= Soy + Vyt + 0.5gt^2
0= =2.37m + 0*t + 0.5(-9.8)t^2
-2.37= -4.9t^2
t= 0.695
0= 2.37 + 0*t + 0.5 (-9.8)t^2
t= 0.695s

~Then again after thinking about this it doesn't matter how fast you serve the ball b/c that only affects the motion in the x direction and the time the ball stays in the air is the same so the commentator is incorrect in physics principles.

You are absolutely correct - It doesn't matter if X component of velocity is 1 mile / hour or 121294812849489175092 miles / hour, since in projectile motion, X component and Y component of motions are independent. Notice the time you found in this question is exactly the same as the previous questions.

 

[~christina~]

Thanks Alot!

 

[shuh]

This is really impressive.. Can we have it as one of the best homework threads?