(Conceptual Problem) Whether the ball can pass through the basket

[daphnelee-mh]

Homework Statement

attached below

Relevant Equations

x=(v1)xt
y=(v1)y-1/2gt^2

Is it proved that the ball can pass through the basket if the positive value of time can be found? I am sure whether I did correct or not.

 

[haruspex]

Your equation at the bottom of the first page (y=...) makes no sense (there is a y on the right which should be t), and its solution as t= would involve square roots.
Write the correct equation and use the t= you obtained from horizontal constraints to substitute for t.

 

[daphnelee-mh]

Your equation at the bottom of the first page (y=...) makes no sense (there is a y on the right which should be t), and its solution as t= would involve square roots.
Write the correct equation and use the t= you obtained from horizontal constraints to substitute for t.

Thanks, I corrected it but how can I prove will it pass through the basket or not?

 

[archaic]

Thanks, I corrected it but how can I prove will it pass through the basket or not?

If the ball's position is given by $(x(t), y(t))$ for a given $t$, and the basket's position is $(a,b)$, then how would you know whether the ball will pass through it or not?

 

[haruspex]

View attachment 266295
Thanks, I corrected it but how can I prove will it pass through the basket or not?

You seem to have substituted $\frac {\sqrt 3}2$ for $v_y$. Should be $v\frac {\sqrt 3}2$.
My instruction was to use the expression you got for t in (1) to substitute for t in your y= equation. By eliminating t, you get a time-independent equation for the trajectory.

 

[aspodkfpo]

View attachment 266295
Thanks, I corrected it but how can I prove will it pass through the basket or not?

If at a certain time, the ball is at the right position both with respect to the x-coordinate and the y-coordinate, the ball goes through the net. If not, the motion forms a parabola which misses the hoop.

 

[daphnelee-mh]

You seem to have substituted $\frac {\sqrt 3}2$ for $v_y$. Should be $v\frac {\sqrt 3}2$.
My instruction was to use the expression you got for t in (1) to substitute for t in your y= equation. By eliminating t, you get a time-independent equation for the trajectory.

I eliminated the t and got the initial velocity at the end. But I have no idea what to do the next to prove it.

 

[haruspex]

I eliminated the t and got the initial velocity at the end. But I have no idea what to do the next to prove it.

You are not using enough information from the picture. If you fix the start and end points of the trajectory and the launch angle, as you have done, there will always be a certain launch velocity which meets the criteria.
You have not said how you determined the launch angle. My reading is that you are supposed to use the angle of the arms. Is that what you did? This means the position of the ball in the picture is additional information. In geometric terms, the question is whether there is a parabola with a vertical axis which passes through a given three points (shoulders, ball in picture, hoop) and at the shoulders has the same slope as the arms.
Indeed, you can treat the problem as purely that and not consider ballistic equations at all.

 

[daphnelee-mh]

You are not using enough information from the picture. If you fix the start and end points of the trajectory and the launch angle, as you have done, there will always be a certain launch velocity which meets the criteria.
You have not said how you determined the launch angle. My reading is that you are supposed to use the angle of the arms. Is that what you did? This means the position of the ball in the picture is additional information. In geometric terms, the question is whether there is a parabola with a vertical axis which passes through a given three points (shoulders, ball in picture, hoop) and at the shoulders has the same slope as the arms.
Indeed, you can treat the problem as purely that and not consider ballistic equations at all.

Sorry but I am not fully understand, I found some mistakes from my work before and corrected as below
I understood that if the starting point is fixed, certain launch velocity can be found therefore I am confused and thinking whether there is another way to prove whether will the ball pass through the basket. The projectile motion equations come first in my mind.

 

[haruspex]

Sorry but I am not fully understand, I found some mistakes from my work before and corrected as below
I understood that if the starting point is fixed, certain launch velocity can be found therefore I am confused and thinking whether there is another way to prove whether will the ball pass through the basket. The projectile motion equations come first in my mind.
View attachment 266333

Ok, you can do it this way, but, as I wrote, you need to use the position shown for the ball. Would the ball, thrown at the angle and velocity you found, pass through that position?
Of course, because you have had to make estimates from a picture, there is going to be some uncertainty in your numbers. Ideally you would track all those uncertainties and end up with a range of possible trajectories for the ball and see whether the shown position lies within them.

 

[aspodkfpo]

Ok, you can do it this way, but, as I wrote, you need to use the position shown for the ball. Would the ball, thrown at the angle and velocity you found, pass through that position?
Of course, because you have had to make estimates from a picture, there is going to be some uncertainty in your numbers. Ideally you would track all those uncertainties and end up with a range of possible trajectories for the ball and see whether the shown position lies within them.

Sorry but I am not fully understand, I found some mistakes from my work before and corrected as below
I understood that if the starting point is fixed, certain launch velocity can be found therefore I am confused and thinking whether there is another way to prove whether will the ball pass through the basket. The projectile motion equations come first in my mind.
View attachment 266333

No, this doesn't work. You have calculated the ball hitting the rim on a path underneath the hoop, rather than reaching a vertex before going back down through the hoop.

 

[aspodkfpo]

No, this doesn't work. You have calculated the ball hitting the rim on a path underneath the hoop, rather than reaching a vertex before going back down through the hoop.

What the heck, I keep getting no solution. Can someone actually solve this question lol.

I keep arriving at this equation which apparently has no solution,

Method:
Not bothering with the use of feet to calculate, scale it after the answer, use the mm.

x: vcos(theta) t(1) + vcos(theta) t (2) = 40

t(1): vsin(theta)/g, from v = u + at

y: (v^2)(sin^2 (theta))/2g - g ((t2)^2)/2 =5, distance to vertex - distance down

t2 from above.

sub in everything into the first equation, v has no solution.

 

[haruspex]

sub in everything into the first equation, v has no solution.

I just noticed @daphnelee-mh is using distances in feet but g in m/s².

 

[aspodkfpo]

I just noticed you are using distances in feet but g in m/s².

Oh, LOL.

 

[aspodkfpo]

I just noticed you are using distances in feet but g in m/s².

Still getting no solution.

 

[haruspex]

Still getting no solution.

Sorry, my previous comment was intended for @daphnelee-mh , but maybe you inherited the same error.
Indeed, you were using mm as measured from the picture. Did you scale g accordingly?

You don't need to break it into two stages. The equation at the end of post #9 (even when corrected) gives a unique positive value for v. Yes, in principle the solution should be checked as having a negative vertical velocity on reaching the basket, but looking at the picture, if the measurements have been accurately transcribed then it is pretty clear the ball cannot end up going upwards through the it.
What made you conclude such was the solution found?

 

[aspodkfpo]

Sorry, my previous comment was intended for @daphnelee-mh , but maybe you inherited the same error.
Indeed, you were using mm as measured from the picture. Did you scale g accordingly?

You don't need to break it into two stages. The equation at the end of post #9 (even when corrected) gives a unique positive value for v. Yes, in principle the solution should be checked as having a negative vertical velocity on reaching the basket, but looking at the picture, if the measurements have been accurately transcribed then it is pretty clear the ball cannot end up going upwards through the it.
What made you conclude such was the solution found?

Nevermind, her answer is right. Breaking it into two also works, but the working out is atrocious. I was thinking about a scenario where theta isn't given, and you solve for theta, hence taking the vertex value.

v=8.25m/s for the quesiton.

Now extremely confused about why symbolab algebra calculator does this wrong. Oh, the calc took it as radians.

 

[daphnelee-mh]

Nevermind, her answer is right. Breaking it into two also works, but the working out is atrocious. I was thinking about a scenario where theta isn't given, and you solve for theta, hence taking the vertex value.

v=8.25m/s for the quesiton.

Now extremely confused about why symbolab algebra calculator does this wrong. Oh, the calc took it as radians.

How you got V=8.25m/s

 

[haruspex]

How you got V=8.25m/s

Did you read post #13? What do you get for v now?

 

[daphnelee-mh]

Did you read post #13? What do you get for v now?

I found v=11.12 ft/s after using g=32.2ft/s^2, but isn't this is the initial velocity required to pass the basket? Seems didn't prove it will pass the basket or not.

 

[haruspex]

I found v=11.12 ft/s after using g=32.2ft/s^2, but isn't this is the initial velocity required to pass the basket? Seems didn't prove it will pass the basket or not.

As I keep telling you, to answer the question you have to make use of the position of the ball in the picture. At the launch angle and velocity you have found, does the ball go through there?

 

[aspodkfpo]

View attachment 266504
I found v=11.12 ft/s after using g=32.2ft/s^2, but isn't this is the initial velocity required to pass the basket? Seems didn't prove it will pass the basket or not.

You did the operations on your calculator wrong for the last question.
Basically you can find the vertex of the parabola. If that vertex is further away than the basket, then it doesn't go down through the basket.

 

[haruspex]

You did the operations on your calculator wrong for the last question.
Basically you can find the vertex of the parabola. If that vertex is further away than the basket, then it doesn't go down through the basket.

I'm not sure whether you are making one point or two.

The algebra was quite correct for finding the launch velocity for which the ball would hit the target, given the launch angle. If "the operations on your calculator wrong" then the numerical answer was wrong. Is that what you mean? I have not checked it.

We already discussed whether it might fail by hitting the underside. Yes, @daphnelee-mh's calculation does not rule that out, but one glance at the picture should be enough to see that will not happen.

The more useful avenue is determining from the picture that the necessary speed has been employed by the thrower. See post #21.

 

[aspodkfpo]

I'm not sure whether you are making one point or two.

The algebra was quite correct for finding the launch velocity for which the ball would hit the target, given the launch angle. If "the operations on your calculator wrong" then the numerical answer was wrong. Is that what you mean? I have not checked it.

We already discussed whether it might fail by hitting the underside. Yes, @daphnelee-mh's calculation does not rule that out, but one glance at the picture should be enough to see that will not happen.

The more useful avenue is determining from the picture that the necessary speed has been employed by the thrower. See post #21.

Two points, 1. Calculator/algebra for her answer in the final line was done wrong.
2. A way for testing whether it falls down and through is, finding the x-coordinate of the vertex of the parabola. If that vertex is further away than the basket, then it doesn't go down through the basket.

 

[haruspex]

Calculator/algebra for her answer in the final line was done wrong.

Yes, I get 27f/s. Is that what you get?

A way for testing whether it falls down and through is, finding the x-coordinate of the vertex of the parabola. If that vertex is further away than the basket, then it doesn't go down through the basket.

Yes, that works. Or find the vertical velocity at the basket, which is easy having found v.

Anyway, as I mentioned in post #8, the simplest approach to the question is purely geometric: is there a parabola with a vertical axis that passes through the three points and has the given tangent at the launch point?

 

[aspodkfpo]

Yes, I get 27f/s. Is that what you get?

Yes, that works. Or find the vertical velocity at the basket, which is easy having found v.

Anyway, as I mentioned in post #8, the simplest approach to the question is purely geometric: is there a parabola with a vertical axis that passes through the three points and has the given tangent at the launch point?

How exactly does your purely geometric method work, is it literally observation?

 

[haruspex]

How exactly does your purely geometric method work, is it literally observation?

Whichever method you use, you have to find the coordinates of three points: the man's shoulder, the ball and the basket. You also need to estimate the launch angle from the angle of the man's arms.
We know that the trajectory is a parabola with a vertical axis, so is representable as $y=Ax^2+Bx+C$. The three points suffice to fix those coefficients, so if the tangent at the shoulder point also matches the launch angle then we know the basket will be hit.