Calculate Initial Acceleration Needed to Throw a Ball from Point A to B

  • Thread starter Thread starter Bosduif
  • Start date Start date
  • Tags Tags
    Curve
AI Thread Summary
To calculate the initial speed needed to throw a ball from point A to point B, the discussion clarifies the importance of the vertical height difference and the angle of throw. The thrower needs to consider both horizontal and vertical components of velocity, particularly at a 45-degree angle where horizontal and vertical velocities are equal. The problem can be simplified by redefining the coordinates to focus on the distance between A and B, while assuming the vertical component is significant. The conversation emphasizes that understanding the acceleration and velocity vectors is crucial for determining the correct initial speed to hit the moving target.
Bosduif
Messages
3
Reaction score
0
Hello

Homework Statement


I have two known points A and B. I want to throw a ball from A to B with a certain angle. Obviously I also know the gravity.

For example: (in metres)

Known
point A: (0, 0, 0)
point B: (30, 0, 10)
direction: (0.949, 0, 0.316)
gravity: 9.82 m/s
angle: 45° or PI / 4

Wanted
acceleration: ?

How can I calculate the initial acceleration needed at point A for the ball to land on point B?

Thanks in advance.
 
Physics news on Phys.org
Do you mean initial velocity?

If you do mean initial acceleration, then how is that acceleration assumed to be applied to the ball, i.e., is it constant acceleration for some time window, or what?
 
Yes, I meant velocity, the initial speed at which the ball is thrown
I'm sorry, English isn't my primary language and I always get the two mixed up
 
Bosduif said:
Yes, I meant velocity, the initial speed at which the ball is thrown
I'm sorry, English isn't my primary language and I always get the two mixed up

No problem.

I would like to clarify one more thing. Are the (x,y,z) coordinates oriented such that +z = up, i.e. point B is 10 meters higher than point A?
 
Oh, I'm using a Y is up system but the position doesn't really matter.
The final purpose is that I have a thrower and a target which moves around. The thrower throws a ball with a certain speed towards the target. The ball always has to hit the target.

In one case know the starting point, the point of destination and the angle (direction vector) and I need to know the speed to multiple with the direction so that it will hit the target.
 
Bosduif said:
Oh, I'm using a Y is up system but the position doesn't really matter.

Well, the vertical coordinate does matter. If the points are at different heights then certain throwing angles might not even work, no matter what velocity you use.

But let's assume both points are at the same height, and we don't even need to work in three dimensions. Just use a coordinate that points from A to B. All that matters is their distance apart. Furthermore, we can define the origin to be point A.

So let's reformulate the problem as follows:

I want to throw a ball from the origin (0,0) to some point (d,0), where d is a horizontal distance. The second (y) component is height.

The angle is constrained to be 45 degrees. Choose an initial velocity so that the ball will hit the target.

We can define the velocity as a function of time as follows.

v(t) = [v_x(t), v_y(t)]

where v_x(t) is the horizontal velocity and v_y(t) is the vertical velocity. We will assume the ball is thrown at t = 0.

Note that the 45 degree throwing angle means

v_x(0) = v_y(0)

We can also define the position as a function of time:

p(t) = [p_x(t), p_y(t)]

where we assume

p_x(0) = p_y(0) = 0

The velocity and position vectors are related as follows

\frac{d}{dt}p(t) = v(t)

Finally, we can define an acceleration vector

a(t) = [a_x(t), a_y(t)]

which relates to velocity as follows

\frac{d}{dt} v(t) = a(t)

So let's start with acceleration. What are a_x(t) and a_y(t)?
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top