Oblique soccer shot calculation task

[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.

 

[johanMT]

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.

I have the impression that you expect me to know physics very well, I'm not a physicist and I know the basics, but I have to practice to pass it in college.

 

[berkeman]

I have the impression that you expect me to know physics very well, I'm not a physicist and I know the basics, but I have to practice to pass it in college.

Okay, but when you say this:

Ok, I see that you didn't say anything specific or help, so let me explain how it should have been done.

That is pretty insulting to the folks who have been trying to help you. Likely it's just a language barrier issue, so we'll let it go at this point.

We are definitely here to help as much as we can, as long as the student shows good effort. BTW, it would help in your future posts if you could review the "LaTeX Guide" link below the Edit window, so that you can post math equations in LaTeX going forward. It makes the math so much easier to read. Thanks

 

[berkeman]

Here are some more tips on using LaTeX to post your math equations:

To post math equations, it's best to use the LaTeX engine that PF provides. There is a helpful "LaTeX Guide" link below the Edit window to get you started. Note that you put double-$ delimiters at the start and end of each stand-alone line of LaTeX, and double-# delimiters at the start and end of in-line LaTeX that does not need to be on its own line. Also, if you right-click on a LaTeX equation in a post, you get a pop-up menu to let you view the LaTeX source or view it in other formats.

LaTeX isn't supported in thread titles, so you can use simple text math in titles if you want.

Note also that PF uses a feature called "lazy LaTeX rendering" that speeds up page loads. When you first post your LaTeX in a thread, you will not see it rendered that first time. Just refresh your browser page to force it to be rendered, and then it should render whenever you come back to that page/thread in the future.

Let me know if you have more questions.

 

[haruspex]

I have the impression that you expect me to know physics very well

Not at all, which makes it all the more important that we responders focus on clearing up the most basic misunderstandings.
The key one in this thread was confusing the initial velocity before the ball is kicked (0) with the initial velocity for the purposes of the SUVAT equations, v. Other thread creators on this forum have made similar mistakes.
There are two lessons to draw from it:

1. The SUVAT equations only apply over a time period during which acceleration is constant. Many processes in the real world involve multiple steps in which the acceleration is only constant during each step, not from one step to the next. In the present case, there are three steps:
   1. The ball is kicked, producing a very high unknown, and maybe inconstant acceleration for a brief time.
   2. The ball flies through the air under constant acceleration due to gravity (since we are ignoring air resistance).
   3. The ball lands, undergoing another very high unknown, and maybe inconstant acceleration for a brief time.
2. This is not so much a matter of knowing physics as of using the intuitive grasp of physics we all have from infancy. Students who think they are not good at physics often mistakenly suppose that getting good at it involves learning lots of formulae and rules. Blindly plugging numbers into rote formulae without imagining the process leads to blunders.