Homework Help: Calculation to launch a golf ball

1. Mar 3, 2015

Maricalue

Hi,

I need your help. Is my calculation correct?

1. The problem statement, all variables and given/known data

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.

2. Relevant 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

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

3. 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)

2. Mar 3, 2015

Staff: Mentor

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.

3. Mar 3, 2015

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

4. Mar 3, 2015

Staff: Mentor

Okay. And sin(45) is 1/√2, so...

5. Mar 3, 2015

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??

6. Mar 3, 2015

Staff: Mentor

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