Calculation to launch a golf ball

[Maricalue]

Hi,

I need your help. Is my calculation correct?


1. Homework Statement
We have to build a launcher that will be able to reach a target that is place anywhere in a prism 3m X 3m X 1m. The launcher have to used a golf ball and gravity as source of energy.

I am trying to calculate what we need as energy. :

-We are thinking about using a pendulum.

-We want it to hit the ball in a tube that we could control to adjust the angle of launching.​

We know :

-that we have to used 45 degrees to maximize our shot.

-that the ball golf as a coefficient of restitution of 0,83 and a mass of 45,9 g.

-that the longest distance we will have to reach will be at 1,5 m to the left or the right, 1 m from the ground and at 3 m from the launcher.​

We think of using a mass of 100g for our pendulum.

Homework Equations

By Pythagoras's theorem, we can calculate the distance we want to reach.

c^2 = a^2 + b^2

Using cinematics equations, I have this :

x=V_ {xo} * t

y=V_ {yo}* t - \frac{1}{2} g t^2

The equation for a inelastic collision is :

V_ {2}=\frac{C_r m_1 V_1 + m_1 V_1}{m_1 + m_2}

For the pendulum, we have the equations of the Law of the Conservation of energy

\frac{1}{2}mV^2= m g h

The Attempt at a Solution

If we have an angle of 45 degrees, we know that the speed in x and in y will be the same and will be equal to :

V_ {xo}= V_ {yo} = V sin 45

By Pythagora's, we know that we need to reach 3,35 m in X and 1 m in Y.

With the equations of cinematics, I've got :

V=\frac{x}{\sqrt{(\frac{2(x-y)}{g})}}

and that give me V= 3,42 m/s

After that, as I know that the ball golf as a coefficient of restitution of C_r = 0.83, I have calculate that I need a speed of V= 4.2896 m/s with the equation of inelastic collision.

Finally, with the equations of the pendulum, I have calculate that I have to have a height of h = 0.937 m.Is that correct?
Thank you!
(Sorry if I have made any english errors)

 

[gneill]

With the equations of cinematics, I've got :

V=x(2(x−y)g)√V=\frac{x}{\sqrt{(\frac{2(x-y)}{g})}}

and that give me V=3,42m/sV= 3,42 m/s

Where does the "2" come from in the 2(x - y) term inside square root? Can you show your derivation?

Note that for 45°, sin(45) = cos(45) = 1/√2.

 

[Maricalue]

It came from the isolation:
1- x=V_ {xo} * t

2- y=V_ {yo}* t - \frac{1}{2} g t^2

In the second equation, I replace t by t=\frac{x}{V_ {xo}} and I replace V_ {xo} =V_ {yo}=Vsin45

I now have :
y=\frac{Vsin45}{Vsin45}*x - \frac{1}{2} g (\frac{x}{Vsin45})^2
And I isolate
y=x - \frac{1}{2} g (\frac{x}{Vsin45})^2
\frac{1}{2} g (\frac{x}{Vsin45})^2 = x-y
(\frac{x}{Vsin45})^2 = \frac{2(x-y)}{g}
\frac{x}{Vsin45}) = \sqrt{\frac{2(x-y)}{g}}
Vsin45 = \frac{x}{\sqrt{\frac{2(x-y)}{g}}}

And now I realized I have forgotten the sin 45 so :
V = \frac{x}{\sqrt{\frac{2(x-y)}{g}}} * \frac{1}{sin45}​

 

[gneill]

Okay. And sin(45) is 1/√2, so...

 

[Maricalue]

So I would need 6,84 m/s.
Then, because of the inelastic collision, I would have to have a speed of 8,57 m/s for my pendulum and a height of 3,7m??

 

[gneill]

So I would need 6,84 m/s.

That looks good.

Then, because of the inelastic collision, I would have to have a speed of 8,57 m/s for my pendulum and a height of 3,7m??

Hmm. I did the calculation quickly, but I didn't see a value that large: Something between 5 and 6 m/s.