Projectile motion of home run ball

[mlbuxbaum]

Homework Statement

major league baseball outfield fences are typically 350 feet away from home plate. Home run balls need to be at least 12 feet off the ground to clear the fence. Calculate the minimun required initial velocity( in mph ) for a home run, for baseballs hit at 15°, 30°, 45°, and 60°. Also given was the height at which the ball was hit from, 3 feet off the gound. Gravity is taken into consideration, 32.2 feet/sec^2.
I need to put the above information into an Excel worksheet and generate a chart for each of the angles above, and i need a formula i can grasp. The teacher gave us one but it just confuses me.

Homework Equations

v_x=v₀costheta₀
X=v₀tcostheta₀
t=x/v₀costheta₀
v_y=v₀sintheta₀-gt
y=y₀+xtantheta₀-1/2g(x²/v₀²cos²theta₀)

The Attempt at a Solution

I don't know if i entered the formulas correctly above or not using the sub and sup scripts forms. The only knowns i have are x is initially 0 and y₀is initially 3. I have got provide 11 points on an x y chart and plot the graphs. I need help in figuring out the equation inputs and the equation itself. I really hope this makes sense to someone.

 

[tiny-tim]

hi mlbuxbaum!

i'm not really grasping what the difficulty is …

you've been given the equations …

X=v₀tcostheta₀
y=y₀+xtantheta₀-1/2g(x²/v₀²cos²theta₀)

… can't you just plug the given angles in, and draw the charts?

 

[mlbuxbaum]

hi mlbuxbaum!

i'm not really grasping what the difficulty is …

you've been given the equations …


… can't you just plug the given angles in, and draw the charts?

I don't know v₀?
Am i correct in assuming that the only values of x&y i have are (0,3) and (350, 12)? and that i plug those values into the equations to get the quadratic? Sorry if i am frustarting you. I am trying really.
To find V₀ would i divide both sides (the x equation) by t(cos$$\theta$$) to get what V₀ is? with t being my x coordinates? 0 to 350?

 

[tiny-tim]

To find V₀ would i divide both sides (the x equation) by t(cos$$\theta$$) to get what V₀ is? with t being my x coordinates? 0 to 350?

i'd put in slightly differently …

you have two simultaneous equations in t,

so you find a formula for t from the x equation, and substitute that for t in the y equation

 

[mlbuxbaum]

i'd put in slightly differently …

you have two simultaneous equations in t,

so you find a formula for t from the x equation, and substitute that for t in the y equation

t=x/v₀cos$$\theta$$₀

i then substitute that value for t into the equation
y=y₀+xtan$$\theta$$₀-.5g(x²/v₀²cos²$$\theta$$)

??

 

[tiny-tim]

(have a theta: θ )

that's right!

so what do you get? ​

 

[ehild]

So you have got the final equation for the x,y coordinates along the path of the projectile.

y=y₀+xtanθ₀-1/2g(x²/v₀²cos²θ₀)

The ball must be at least 12 feet hight at 350 feet distance from the home plate (if only I knew what the home plate is... )So it must pass the point (12;350). Choose a θ₀ and plug in x=350, y=12, and solve the equation for v₀. You can make an excel table with θ₀, v₀ and plot v₀ as function of θ₀. I think you even can make Excel compute v₀ for each θ₀ if you express v₀ in terms of x,y and θ₀.

ehild
.

 

[mlbuxbaum]

(have a theta: θ )

that's right!

so what do you get? ​

y=3+x(.2679)-16.1(x²/v₀²(.9330))

using 32.2 ft/sec² for g

 

[tiny-tim]

(have you missed out an x ?)

the object is to find v₀ …

what do you get for v₀ ?​

 

[mlbuxbaum]

(have you missed out an x ?)

the object is to find v₀ …

what do you get for v₀ ?​

i fixed the missing x.
Finding v₀ is where i am having my issue. I'm not sure how to get it alone

 

[mlbuxbaum]

y=3+x(.2679)-16.1(x²/v₀²(.9330))

using 32.2 ft/sec² for g

12=3+.2679x-(16.1x²/.9330v₀²

.9330(12-3-.2679x/-16x²)=v₀²

$$\sqrt{}$$.9330(12-3-.2679x/-16x²)=$$\sqrt{}$$v₀²

v₀=$$\sqrt{}$$.9330(12-3-.2679x/-16x²)

 

[ehild]

Plug in the value given for x (x=350) and check the parentheses.

"12=3+.2679x-(16.1x²/.9330v₀²" is wrong.

Correct: 12=3+0.2679-16.1x²/(0.9330v₀²)

Continue from here.

ehild

 

[mlbuxbaum]

Plug in the value given for x (x=350) and check the parentheses.

"12=3+.2679x-(16.1x²/.9330v₀²" is wrong.

Correct: 12=3+0.2679-16.1x²/(0.9330v₀²)

Continue from here.


ehild[/QUOTE
Did you forget the x after class .2679?

 

[ehild]

ehild[/QUOTE
Did you forget the x after class .2679?

ops! yes. So it is 12=3+0.2679x-16.1x²/(0.9330v₀²)

 

[mlbuxbaum]

i went to the tutor on campus yesterday and he walked me through the problem. The tutor had the same class, same instructor, same problem a few years ago.

12=3+(v₀sin15)(350/v₀cos15)-.5(32.2)(350²/v₀²cos²15)

so plugging numbers in and working it through

9=350tan15-16.1(131295.1043/v₀²)

working it through again...

-84.78= -2113851.179/v₀²

to get v₀² alone.


-2113851.179/-84.78 = v₀²

v₀=$$\sqrt{}$$-2113851.179/-84.78

v₀=157.9 feet/sec

now i need to figure out how to get the x values needed via the quadratic equation

 

[ehild]

x is the distance of the fence from the home plate. It is 350 feet. Just use the equation the teacher gave you

9=350 tan(theta0)-16.1*122500/(v₀²cos²theta0) and plug in the angles one by one and calculate v0 for each angle.

ehild

 

[mlbuxbaum]

x is the distance of the fence from the home plate. It is 350 feet. Just use the equation the teacher gave you

9=350 tan(theta0)-16.1*122500/(v₀²cos²theta0) and plug in the angles one by one and calculate v0 for each angle.

ehild

I worked the V₀ for 15 degrees to be 157.9030425
30 degrees at 116.7051809 45 degrees at 107.5520743 and 60 degrees at 114.9329655

now i need to find out how to find the eleven (x,y) coordinants i need to use so i can plot a curve. I am inputting the formulas into my excel worksheet and what would i use for 't' ?

 

[ehild]

Why do you need the time? You have to make a vo(theta) plot, don't you? Vo can be expressed in terms of theta and input to your Excel worksheet and calculated for as many theta as you like. For example, you can start from 10° and step by 5 degrees up to 60°.

The expression for vo is: vo=1404.37/[cos(theta)√(350 tan(theta)-9)]. Input this equation into Excel.
Anyway, your results for vo at 15, 30, 45, and 60 degrees were correct. Good job!

ehild

 

[mlbuxbaum]

thank you for your help. I inserted the equation into my x,y chart in the y column. I started my x column at 0 and y column at 3. I need 11 x,y points for my chart (per the instructor) so i divided 350 by 11, and each division workes out to be ~31.82. So now i have my x value to plug into the equation. I can try and post my final x,y charts for 15,30,45 and 60 degrees and their respective plots. Not sure how to do that but i'll try.

Thanks to you and Tiny Tim for all the help.

 

[mlbuxbaum]

Why do you need the time? You have to make a vo(theta) plot, don't you? Vo can be expressed in terms of theta and input to your Excel worksheet and calculated for as many theta as you like. For example, you can start from 10° and step by 5 degrees up to 60°.

The expression for vo is: vo=1404.37/[cos(theta)√(350 tan(theta)-9)]. Input this equation into Excel.
Anyway, your results for vo at 15, 30, 45, and 60 degrees were correct. Good job!

ehild

I used this in the column for my unkown Y value
=y₀+xtan$$\theta$$-.5g(x²/v₀²cos$$\theta$$)

i used my calculated x values and the radian values for 15,30,45, and 60 degrees

 

[mlbuxbaum]

i entered in your formula and it doesn't equal what i got. What is 1404.37?

 

[ehild]

Do you want to plot out the orbit of the ball at the given angles? If so, use x values from 0 to 150 or so, and use vo you calculated for the given angle. There will be four curves and you can compare the orbit of the ball if you throw it at different angles and with different initial speeds.

What I said is how to use Excel to calculate the vo values at given angles. That 1404.37 is the squre root of (g/2 *x²).

ehild

 

[mlbuxbaum]

you are correct, that is what i needed to do. I just wanted to put the formula into excel that would calculate v0.

 

[mlbuxbaum]

I have attached my file.

 

[ehild]

I have attached my file.

Oh, this is simply beautiful. Really. You did it very well!

ehild

 

[mlbuxbaum]

Oh, this is simply beautiful. Really. You did it very well!

ehild

Thank you. If it wasnt for the help of you two guys, i would still be scratching my head. I have some polishing to do on the final copy, but the data is there now. And best of all, i actually learned a few things in the process...cool.

Thanks again guys !