- 20 Nov '13 10:16The next project that I am doing with my 13 year old daughter is to build a catapult.

The experiment is to find the angle of release that lets the projectile travel furthest. Also it is an exercise where I can teach her to use hand tools to build the machine.

The idea is to construct it so the 'firing' arm will begin in the horizontal position and travel through 90 degrees to the vertical, hit a stop and release the projectile. With it mounted on a timber base we can lift one end to alter the elevation in say 5 degree increments to carry out the experiment.

My question to you all is can the answer be found by maths calculation and if so how?

Thanks for reading this - 20 Nov '13 20:15 / 2 editsAnd in order to understand the derivation for why a projectile takes the (vertical) path

y = y0 + Vy0 * t - 1/2 * g * t^2

requires basic calculus:

position = y

velocity = y' = Vy0 - g * t

acceleration = y'' = -g

Although that's pretty advanced stuff for a 13-year-old. I was still doing this stuff in first-year physics as an engineering student. - 20 Nov '13 21:27Thanks to both of you for your comments.

Yes Mr Knight, agreed. The maths is for me for now. My daughter will be making the machine and conducting the practical side of things. She is quite keen to chop up pieces of wood, drill holes and make something.

She has already asked about the strength of the spring which of course lets me introduce her to Hooks law. - 20 Nov '13 23:30 / 1 edit

Actually, that's not quite the proper scenario. The angle which maximizes the distance traveled is 45 deg only when the initial height of the object is zero.*Originally posted by AThousandYoung***To calculate the answer requires some simple trigonometry. It's a classic mechanics (physics) problem.**

http://www.wired.com/wiredscience/2010/09/maximum-range-in-projectile-motion/

Since you have catapult, you should take into account the initial height of the object when it is released.

However, analytically that is a tricky task

From (1)

x = V*cos(A)*t

Where;

x = distance

V = magnitude of initial velocity

A = angular direction of initial velocity

t = time

and (2)

y = h + V*sin(A)*t -1/2*g*t^2

where;

y = height at any time "t"

h = initial height

g = accel. due to gravity

V = magnitude of initial velocity

A = angular direction of initial velocity

t - time

After Sub(1) into (2) and simplifiy to eq (3)

y = h + tan(A)*x -1/2*g*x^2/cos^2(A)

the total distance traveled occurs when y = 0 So,

0 = h + tan(A)*x -1/2*g*x^2/cos^2(A) eq(4)

Then you would have to solve for x(A) (distance "x" as a function of the angle "A" )

Take the first derivative of that x(A) and set it equal to zero and solve

dx/dA = 0

I should note that the last couple steps are a disaster if you are attempting to arrive at an analytic result.

However, you could always make a plot by plugging in various values of "A" between 0 and 90 deg into eq(4), solve the quadratic for x, then graph x vs. A. and locate the maximum "x" and the corresponding "A" from the chart. - 21 Nov '13 00:29Thanks for your input Jo.

Missed that point altogether i.e an offset of the release height above the landing point.

I think the old saying goes 'If it was easy then everyone would do it'

And of course the classic 'A problem is the gift wrapping of a solution'

Next time there is athletics on the TV I will be watching the throwing events with a keen eye.

Cheers - 21 Nov '13 01:35 / 1 edit

One problem with classic projectile physics is that it totally neglects to account for air resistance of any kind.*Originally posted by eltricky*

Next time there is athletics on the TV I will be watching the throwing events with a keen eye.

Many of the Olympic style throwing events involve a significant amount of aerodynamic flight, which changes things considerably as far as optimal trajectories. - 22 Nov '13 02:35

Thats not so much a probem for classical mechanics, as it is a problem of mathematical complexity.*Originally posted by forkedknight***One problem with classic projectile physics is that it totally neglects to account for air resistance of any kind.**

Many of the Olympic style throwing events involve a significant amount of aerodynamic flight, which changes things considerably as far as optimal trajectories.

You have to solve for x(t) from

m*d²x/dt² = - k*(dx/dt)²

and y(t) from

m*d²y/dt² = k*(dy/dt)² -mg

So what to do (unless you can derive the solutions from above) is to say that its not going to go as high, and not as far as what our theory predicts. - 29 Nov '13 21:56 / 1 editLet me try this:

Say we have the Vx and Vy (speeds) of the projectile:

The time in flight merely depends on Vy and initial height H

First work out time to zero Y speed (T_0)

Speed at any time = Initial_velocity + acceleration*time

When the projectile has gone up and back down to its initial height then

it must be moving at the opposite y-speed to its initial speed

since change in speed = acceleration*time

2Vy = Gxt

t = 2Vy/G

(which is why starting and ending at the same height is simple)

However in our case it must still get to the ground, lets say this time is T_1

H = T_1 + Vy.T_1 + G.T_1**2/2

T_1^2 + (2/G)(1+Vy).T_1 = 2H/G

T_1^2 + T1.2(1+Vy)/G + ((1+Vy)/G))^2 = 2H/G + ((1+Vy)/G)^2

( T_1 + (1+Vy)/G ) ^2 = 2H/G + ((1+Vy)/G)^2

T_1 + (1 + Vy)/G = sqrt( 2H/G + ((1+Vy)/G)^2)

T_1 = sqrt( 2H/G + ((1+Vy)/G)^2) - (1 + Vy)/G

so total time in flight = T_1 + t =

sqrt( 2H/G + ((1+Vy)/G)^2) - (1 + Vy)/G + 2Vy/G

and total distance = Vx times that

so

Distance = Vx.( sqrt( 2H/G + ((1+Vy)/G)^2) - (1 + Vy)/G + 2Vy/G )

we know speed_squared (SS) = Vx^2 + Vy^2

so Vx^2 = SS - Vy^2

hmm, gets a bit tricky - 07 Dec '13 02:30 / 1 edit

I should note that eq(4) (which neglects friction as F.K. pointed out) that I derived can be explicitly solved for x(A) from the quadratic formula.*Originally posted by joe shmo***Actually, that's not quite the proper scenario. The angle which maximizes the distance traveled is 45 deg only when the initial height of the object is zero.**

Since you have catapult, you should take into account the initial height of the object when it is released.

However, analytically that is a tricky task

From (1)

x = V*cos(A)*t

Where;

...[text shortened]... for x, then graph x vs. A. and locate the maximum "x" and the corresponding "A" from the chart.

0 = c + b*x + a*x²; where

c = h

b = tan(A)

a = 1/2*g*sec²(A)

so,

x = -tan(A) ± Sqrt(tan²(A) + 2*g*h*sec²(A))/(-g*sec²(A) )

from which you can find dx/dA = 0, and theoretically solve it ( however I suspect a method of approximation would need to be employed).

Anyhow, I pretty sure the topic is dead, but I just wanted to clarify. - 07 Dec '13 02:41

Something isn't quite right here I believe. You seem to be adding time and distance with this equation.*Originally posted by iamatiger***Let me try this:**

Say we have the Vx and Vy (speeds) of the projectile:

The time in flight merely depends on Vy and initial height H

First work out time to zero Y speed (T_0)

Speed at any time = Initial_velocity + acceleration*time

When the projectile has gone up and back down to its initial height then

it must be moving at the opposite y-speed to i ...[text shortened]... y/G )

we know speed_squared (SS) = Vx^2 + Vy^2

so Vx^2 = SS - Vy^2

hmm, gets a bit tricky

"H = T_1 + Vy.T_1 + G.T_1**2/2"

H = time + distance + distance ?? - 08 Dec '13 12:39

Aha! Thank you, yes, you are correct*Originally posted by joe shmo***Something isn't quite right here I believe. You seem to be adding time and distance with this equation.**

"H = T_1 + Vy.T_1 + G.T_1**2/2"

H = time + distance + distance ??

I am always too careless at maths.

So, corrected:

H = Vy.T_1 + G.T_1**2/2

2H/G = (2Vy/G).T_1 + T_1^2

sqrt(2H/G+Vy^2/G^2) = T_1 + Vy/G

T_1 = sqrt(2H/G+Vy^2/G^2) - Vy/G {only want positive root}

giving total time in flight = Vy/G + sqrt(2H/G+Vy^2/G^2)

Distance = Vx(Vy/G + sqrt(2H/G+Vy^2/G^2))

using

Vx^2 = SS - Vy^2

Distance = sqrt(SS - Vy^2 )(Vy/G + sqrt(2H/G+Vy^2/G^2))

so differentiating

d(distance)/d(Vy) =

sqrt(SS - Vy^2 )(1/G + d/dy{ sqrt(2H/G+Vy^2/G^2))}

+ (Vy/G + sqrt(2H/G+Vy^2/G^2))(d/dy{sqrt(SS - Vy^2 )})

from the chain rule

d/dx f(g(x)) = f'(g(x)).g'(x)

so

d/dx sqrt(g(x)) = g'(x)/(g(x)

d(distance)/d(Vy) =

sqrt(SS - Vy^2 )(1/G + (2Vy/G^2)/(2H/G+Vy^2/G^2)

+ (Vy/G + sqrt(2H/G+Vy^2/G^2))(-2Vy/(SS - Vy^2 ))

so we need to solve for

0 =

sqrt(SS - Vy^2 )(1/G + (2Vy/G^2)/(2H/G+Vy^2/G^2)

+ (Vy/G + sqrt(2H/G+Vy^2/G^2))(-2Vy/(SS - Vy^2 )) - 10 Dec '13 00:08

I suppose that if all the math is done correctly it would be a valid result by first solving for V_y, then V_x from*Originally posted by iamatiger***Aha! Thank you, yes, you are correct**

I am always too careless at maths.

So, corrected:

H = Vy.T_1 + G.T_1**2/2

2H/G = (2Vy/G).T_1 + T_1^2

sqrt(2H/G+Vy^2/G^2) = T_1 + Vy/G

T_1 = sqrt(2H/G+Vy^2/G^2) - Vy/G {only want positive root}

giving total time in flight = Vy/G + sqrt(2H/G+Vy^2/G^2)

Distance = Vx(Vy/G + sqrt(2H/G+Vy^2/G^2))

using

Vx^2 ...[text shortened]... t(SS - Vy^2 )(1/G + (2Vy/G^2)/(2H/G+Vy^2/G^2)

+ (Vy/G + sqrt(2H/G+Vy^2/G^2))(-2Vy/(SS - Vy^2 ))

V_x = Sqrt(SS - V_y^2)

and finding the corresponding angle from

tan(Angle) = V_y/V_x

But that final equation would be a absolute nightmare for sure! So I think I'm going to have to stick with my method. Who knows though, seems to be a difficult solution in either case. - 11 Dec '13 00:46ok , thanks Jo

Now we need to simplify:

sqrt(SS - Vy^2 )(1/G + (2Vy/G^2)/(2H/G+Vy^2/G^2))

+ (Vy/G + sqrt(2H/G+Vy^2/G^2))(-2Vy/(SS - Vy^2 ))

= 0

{I replaced an errant bracket}

so multiply through by sqrt(SS - Vy^2 )

(SS - Vy^2 )(1/G + (2Vy/G^2)/(2H/G+Vy^2/G^2))

+ (Vy/G + sqrt(2H/G+Vy^2/G^2))(-2Vy/sqrt(SS - Vy^2 )) = 0

Do some substitutions:

(SS - Vy^2 ) = A

(2H/G+Vy^2/G^2) = B

so

A(1/G + (2Vy/G^2)/B)

+ (Vy/G + sqrt(B))(-2Vy/sqrt(A) = 0

multiply out the rightthand part and move the lefthand part to the other side

2Vy^2/(G.sqrt(A)) + 2Vy.sqrt(B/A) = A(1/G + (2Vy/G^2)/B)

square both sides:

4Vy^4/(G^2.A)+4Vy^2.(B/A) + 8Vy^3.sqrt(B)/G = A^2(1/G + (2Vy/G^2)/B)^2

now we can isolate sqrt(B) on its own, square both sides again, and we at least we have a polynomial without square roots in, but it still looks complex!