Oblique soccer shot calculation task

[johanMT]

Homework Statement

Oblique shoot task

Relevant Equations

Vx = v0 * cos alpha
Vy = v0 * sin alpha - gt

D = v0 * cos alpha* t
Y = v0 * t * sin alpha - 1/2 *g * t^ 2

Task:
A football player hits the ball, which is at rest on the ground, so that it moves at an angle of 30° in relation to the horizontal plane and falls to the ground again after 3 seconds. a) How far from the point of impact will the ball fall? b) Calculate the velocity vector of the ball on the upward trajectory at the moment when it is at half its maximum height relative to the ground.

My solution:

a) The ball will fall to the ground after 3 seconds. As the ball is at rest on the ground before impact, we can assume that the initial height of the ball is zero. Also, the ball moves at an angle of 30° in relation to the horizontal plane. When it's time to determine the horizontal displacement of the ball, we can use the horizontal displacement equation: x = v * t * cos(θ) where x is the horizontal displacement of the ball, v is the speed of the ball, t is the time and θ is the angle. Since the time is 3 seconds and the angle is 30°, we can calculate the horizontal displacement: x = v * 3 * cos(30°) x = v * 3 * sqrt(3) / 2 = 0m

(since cos(30°) = sqrt(3) / 2)

b) To calculate the velocity vector of the ball on the upward trajectory at the moment when it is at half the maximum height, we can use the equation for the vertical displacement: y = v * t * sin(θ) - 1/2 * g * t^2 where y is the vertical displacement of the ball, v is the speed of the ball, t is the time, θ is the angle and g is the acceleration of the earth's gravity. Since the time when the ball is at half its maximum height is equal to half of the total time (t/2), we can calculate the vertical displacement: y/2 = v * t/2 * sin(30°) - 1/2 * g * (t/2)^2 y/2 = v * t/2 * 1/2 - 1/2 * g * (t/2)^2 (since sin(30°) = 1/2) y/2 = v * t/4 - 1/8 * g * t^2 Also, since the ball is at half its maximum height, we can assume that the vertical displacement (y) is zero. 0 = v * t/4 - 1/8 * g * t^2 0 = v/4 * t - 1/8 * g * t^2 From this equation we can express the vector speed on the upward path: v = 1/2 * g * t Substituting the known values of the acceleration of the earth's gravity (g = 9.8 m/s^2) and time (t = 3 / 2 = 1.5 s), we can calculate the velocity vector: v = 1/2 * 9.8 m/s^2 * 1.5 s v = 7.35 m/s

 

[PeroK]

If ##v = 7.35 m/s##, then the ball is going to take a lot less than ##3s## to hit the ground. Even if kicked vertically upwards.

 

[haruspex]

x = v * 3 * cos(30°)

Right, but you have not found v.

Since the time when the ball is at half its maximum height is equal to half of the total time (t/2)

Wrong.

since the ball is at half its maximum height, we can assume that the vertical displacement (y) is zero

Eh? Height is vertical displacement here. How can its maximum be zero?

If ##v = 7.35 m/s##, then the ball is going to take a lot less than ##3s## to hit the ground. Even if kicked vertically upwards.

That's the OP's answer to part b, the velocity when at half maximum height.
But, @johanMT, it asks for the velocity vector.

 

[johanMT]

Right, but you have not found v.

Wrong.

Eh? Height is vertical displacement here. How can its maximum be zero?

That's the OP's answer to part b, the velocity when at half maximum height.
But, @johanMT, it asks for the velocity vector.

In a) Part you should find the initial speed, it just says that the ball is at rest according to that logic it should be that the initial speed is zero, but then how to calculate how far it fell when the formula for range requires knowledge of the initial speed, and from this it turns out to me that it is zero, what would mean that the speed times the cosine of the angle times the time that the distance is 0 meters which makes no sense.

 

[johanMT]

Right, but you have not found v.

Wrong.

Eh? Height is vertical displacement here. How can its maximum be zero?

That's the OP's answer to part b, the velocity when at half maximum height.
But, @johanMT, it asks for the velocity vector.

In part b isn't half of the upward trajectory when time t is half?

 

[haruspex]

according to that logic it should be that the initial speed is zero

The SUVAT equations (##s=ut+\frac 12at^2## etc.) only apply to constant acceleration. The player kicks the ball (first acceleration), then gravity takes over (different acceleration). So you can only apply those equations for the free fall phase after the player has kicked the ball. The initial velocity is then not zero.

 

[haruspex]

In part b isn't half of the upward trajectory when time t is half?

First, half of what time? The given total time is 3 seconds. Where will the ball be at 1.5s?
If you mean half the time to reach maximum height, that would be true for constant vertical velocity; but it isn’t.

 

[johanMT]

First, half of what time? The given total time is 3 seconds. Where will the ball be at 1.5s?
If you mean half the time to reach maximum height, that would be true for constant vertical velocity; but it isn’t.

Ok ,then I really don' t know this part.

 

[johanMT]

The SUVAT equations (##s=ut+\frac 12at^2## etc.) only apply to constant acceleration. The player kicks the ball (first acceleration), then gravity takes over (different acceleration). So you can only apply those equations for the free fall phase after the player has kicked the ball. The initial velocity is then not zero.

I don't have a given path in the task, and I'm looking for speed, which means that I will then have two unknowns in the equation, speed and path, and I only need speed.

 

[berkeman]

Thread closed for Moderation...

 

[berkeman]

Thread is reopened for now. @johanMT -- Do not use AI chatbot text to try to show your work in your OP. That is against the PF rules, especially without attribution to the chatbot.

 

[hutchphd]

Was the entire answer first answer chatbot? It certainly parses oddly to my sensibillities. Perhaps it should be retrospectively stricken. (Chatbot part lined out)

 

[Lnewqban]

 

[hutchphd]

It would be useful policy perhaps to require annotation of AI. Perhaps scarlet letters (where is Hester Prynne when we need her ?) We could all learn perhaps.

 

[hmmm27]

Was the entire answer first answer chatbot?

I'm not sure why anybody expects a chat-AI to do physics. My calculator writes shitty essays, and my toaster doesn't even make a good cup of tea.

I suspect even the equations (with the wrong answers based on wrong suppositions and interpretations of the question) were also AI.

 

[berkeman]

It would be useful policy perhaps to require annotation of AI. Perhaps scarlet letters (where is Hester Prynne when we need her ?) We could all learn perhaps.

The following was added to the PF Rules a couple months ago:

ChatGPT and AI-generated text

• Posting AI-generated text without attribution is categorically disallowed and will lead to a warning and an eventual permanent ban with continued use.
• Answering a science or math question with AI-generated text, even with attribution, is not allowed. AI-generated text apps like ChatGPT are not valid sources.
• Threads about the technology and cultural impact behind AI like ChatGPT are allowed
• Usage of AI-generated text output in entertainment threads in General Discussion with attribution is allowed.

 

[berkeman]

Was the entire answer first answer chatbot?

All of their "work" toward the solution was AI generated (just paste it into ZeroGPT yourself to see that it comes back with a 100% match). It's a new way for students to, um, get help on their homework without actually doing much work or studying.

 

[haruspex]

I don't have a given path in the task, and I'm looking for speed, which means that I will then have two unknowns in the equation, speed and path, and I only need speed.

As @Lnewqban shows in post #13, the path is a parabola, but you do not need to know that. Just use the SUVAT equations correctly.
The ball is kicked with unknown speed v at angle 30° above the horizontal. Write an expression for its vertical velocity, and from that the time it will take to reach maximum height. How does that relate to the given 3s?

 

[phinds]

It's a new way for students to, um, get help on their homework without actually doing much work or studying.

And to think, in 30 years, our kids will be driving over bridges designed by MEs who made it through on the back of chatGPT.

 

[berkeman]

Or maybe going on submersible tours to the Titanic!

Oh wait...

 

[PeroK]

And to think, in 30 years, our kids will be driving over bridges designed by MEs who made it through on the back of chatGPT.

 

[Lnewqban]

@johanMT

The pure vertical movement is independent of the trajectory because is only governed by gravity.
The pure horizontal movement is a constant velocity one (no acceleration), as long as the ball is in the air.
What we see is a simultaneous combination of both movements.

 

[johanMT]

@johanMT

The pure vertical movement is independent of the trajectory because is only governed by gravity.
The pure horizontal movement is a constant velocity one (no acceleration), as long as the ball is in the air.
What we see is a simultaneous combination of both movements.

View attachment 330035

View attachment 330036View attachment 330037

if I hit a ball that is at rest on the ground at an angle of 30, it means that it has no speed in the x component because it is at rest, but it has a speed in the y component because gravity acts on it

 

[PeroK]

if I hit a ball that is at rest on the ground at an angle of 30, it means that it has no speed in the x component because it is at rest, but it has a speed in the y component because gravity acts on it

You seem confused about the concept of kicking a ball.

This is approximately the scenario you are analysing here:

 

[johanMT]

You seem confused about the concept of kicking a ball.

This is approximately the scenario you are analysing here:

Yes I see this, this is Magnuss efect.

 

[PeroK]

Yes I see this, this is Magnuss efect.

I'm not sure about that. We are not talking about the sidespin.

The angle here is less than 30 degrees and the ball lands after less than 2 seconds. But, the critical thing is that the ball does not go straight up in the air! The ball would have landed after about 40m, perhaps.

That gives you a physical scenario to help you get a grip on the problem, hopefully.

 

[PeroK]

That said, this is not the easiest projectile motion problem. You perhaps ought to try some simpler problems first.

 

[johanMT]

That said, this is not the easiest projectile motion problem. You perhaps ought to try some simpler problems first.

This is from the exam, that's why I asked

 

[PeroK]

This is from the exam, that's why I asked

You need to do some studying if you want to do problems like this. ChatGPT isn't much help when it comes to problems that involve calculations. You need a textbook or course material.

 

[haruspex]

if I hit a ball that is at rest on the ground at an angle of 30, it means that it has no speed in the x component because it is at rest, but it has a speed in the y component because gravity acts on it

You don't seem to have read and understood my post #6.
The ball is at rest before it is kicked. The SUVAT equations apply from just after kicking the ball. The "initial velocity" occurs then. It will not be zero.

 

[Lnewqban]

if I hit a ball that is at rest on the ground at an angle of 30, it means that it has no speed in the x component because it is at rest, but it has a speed in the y component because gravity acts on it

Your statement would be correct only if you would replace "speed" with "acceleration".
If you did not mean acceleration, but speed; then, you will need to revisit the basic concepts of velocity and acceleration prior to continue on with this type of problems.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/vel2.html#c1

http://hyperphysics.phy-astr.gsu.edu/hbase/acca.html#c1

If you return to the animations shown in post #22 above, you will see that the ball has velocity (red arrow), which we imaginarily decompose in variable vertical velocity (green arrow) and horizontal constant velocity (blue arrow).

 

[haruspex]

Your statement would be correct only if you would replace "speed" with "acceleration".

The second statement, yes. But the first, regarding the x component, is still wrong because the OP is thinking of the motion before the kick instead of the motion immediately after.

 

[johanMT]

Ok, I see that you didn't say anything specific or help, so let me explain how it should have been done.
So, in the part under a, the ball is on the ground, so its height is zero.

y=0

y=v* sin alpha* t - 1/2*g*t^2=0
from this we derive the initial velocity

v=29.43 m/s

This v insert into the range formula

D=x

X=v * cos alpha* t = 76.46 m

b Part

We are looking for the maximum height, the maximum height is when the speed of the y component is equal to zero, we extract the maximum time from that formula and insert it into the formula for the maximum speed.

Hmax => Vy=0

Vy= v* sin alpha - g*t=0
t=1.5 m/s = t max

Hmax = v* sin alpha*tmax - 1/2*g*tmax^2
Hmax=11.04m
Hmax/2=5.52m

Given that we are looking for half of the maximum height, we divide it by 2. After that, we extract from the height formula the time that is maximum when it is at half of the maximum speed. It is a quadratic equation and we look at the result when t is smaller, which in this case is t1

Hmax/2=v*sin alpha*t - 1/2*g*t^2

t1=0.44 s ✓
t2=2.56 s ×

and we insert that t1 into the formula for the y component of the speed and the x component of the speed remains the same because it is horizontal.

Vy = v* sin alpha - g*t1= 10.3986 m/s
Vx = v * cos alpha= 25.49 m/s
And finally we just write down the vector.

v=25.59 i^ + 10.3986 j^

That's it.

 

[haruspex]

y=v* sin alpha* t - 1/2*g*t^2=0
from this we derive the initial velocity

How? What numbers are you plugging in for t and y, and on what reasoning?
I am not saying your method is incorrect, just that you have not explained it adequately.

I note this is much more reasonable than your initial attempt. Is it your own work or has someone else provided this solution? Do you now understand what was wrong with your post #1 attempt?

I see that you didn't say anything specific or help

Is that directed at a particular response or to the responses you've had on this thread in general? As a responder, my top priority is to help you understand where you went wrong. That will be the most helpful to you in future. Getting the answer to the specific question is secondary.

 

[johanMT]

How? What numbers are you plugging in for t and y, and on what reasoning?
I am not saying your method is incorrect, just that you have not explained it adequately.

I note this is much more reasonable than your initial attempt. Is it your own work or has someone else provided this solution? Do you now understand what was wrong with your post #1 attempt?

Is that directed at a particular response or to the responses you've had on this thread in general? As a responder, my top priority is to help you understand where you went wrong. That will be the most helpful to you in future. Getting the answer to the specific question is secondary.

When we hit the ball, we hit it off the floor and it travels along a path and ends up on the ground again. y denotes the height, but since there is no height, the conclusion is that the height is zero, so we assign the value zero to y. For t, we include the fall time, which is 3 seconds as stated in the task. So, the formula is y=v*sin alpha*t - 1/2*g*t^2 when we include it is actually: 0=v*sin30°*3s - 1/2 *9.81m/s^2*3^2s^2 And from that we derive v.