What is maximum range of the catapult?

Question

The arm of the catapult is R=10ft long,
http://www.knightsedge.com/medieval-weapons/medieval-catapult-4804.jpg
and starting from initial horizontal position it acclerates projectile at the rate s=100 rad/s².
The crew can regulate the angle at which projectile is relesead.

g = 32 ft/s², no air resistance

The size of the catapult is negligibly small compared to projectile range.
(It's clear from the data provided, but I want to state this assumption explicitly).

Answer 1

α = 100 rad/s^2
= dω/dt
= ωdω/dθ

ω^2 = (2e2)θ
= v^2 / 100
v^2 = (2e4)θ

This is the angle of hte arm upon release, so the projectile is fired at an angle of elevation of π/2 - θ

Range, R = v^2 / g * sin(2θ)
= (2e4/g) * θ * sin(2θ)

Now we need to find θ which maximizes R

dR/dθ = (2e4/g) * [sin(2θ) + 2θ*cos(2θ) ] = 0

We can solve this numerically to get
θ = 1.014 radians = 58.12 deg
R_max = 568.7 feet

Answer 2

Catapult Range

Answer 3

Convert the acceleration to feet. The arm is 10 feet in length so the rads/s^2 can be multiplied by this to get a f/s^2.

a = 10*100 rads/s^2 = 1000 f/s^2

Now: s = (1/2)a t^2
For s we will use the arc distance and that is 10A where A (in radians) is the angle the arm is rotated through before the projectile is released.

10 A = (1/2)(1000)t^2
t = SQRT(A/50)

Also: v = at (assuming constant acceleration)
v = 1000SQRT(A/50)

We now need the time of flight T. This is twice the time it takes the projecticle velocity to go from it's initial veritcal velocity to 0 (the maximum point of the trajectory). If A is the angle the arm is rotated through then pi/2 - A is the angle at which the projecticle is fired so:
vSIN(pi/2 - A) = gT/2
1000SQRT(A/50)COS(A) = 16T
T = 62.5 SQRT(A/50)COS(A)

If vh is the horizontal velocity then vh = vCOS(pi/2 - A):
Horizontal distance = vhT = vCOS(pi/2 - A)T = D
D= 1000SQRT(A/50)COS(pi/2 - A)* 62.5 SQRT(A/50)COS(A)
D = 1000*62.5 (A/50)SIN(A)COS(A)
D = (1250A)SIN(A)COS(A)

dD/dA = 0 = (1250A)[COS^2(A) - SIN^2(A)] + 1250SIN(A)COS(A)

(A)[COS^2(A) - SIN^2(A)] + SIN(A)COS(A) = 0
A = SIN(A)COS(A) / [SIN^2(A) - COS^2(A)]

A = about 58.12 degrees = about 1.0144 radians

Distance = (1250A)SIN(A)COS(A) = 569 feet

Answer 4

No you wouldn't. You need to figure out the relationship between time spent accelerating vs. the angle of release. Since typically you could say yhe best angle with no air resistance is 45degrees, that would be a good starting point, but its not necessarily valid here since if you waited a bit longer it would be traverling faster. Though at some point the acceleration will have less of an affect that the added angle above the horizontal plane. I just dont want to do the damn math.

Answer 5

About $100...

Answer 6

Find the angle which provides the longest horizontal
distance.

Maximize the horizontal distance:
d=Vo*t ; where Vo is the horizontal velocity and t is the
time, and d is the horizontal distance

Vo=velocity*cos(theta)

Convert from linear to angular:
s=100rad/s^2, R=10ft
a=100*10 ft/s^2
=1000 ft/s^2

Vo=(a*t)*cos(theta)
theta=arccos(Vo/(a*t))

Vv=(a*t)*sin(theta)

d=[(a*t)*cos(theta)]*t
d=[(1000*t)*cos(theta)]*t
d=1000*t^2*cos(theta)

Maximimize the distance by taking the derivative and
solving for 0, but need to solve implicitly

The position function of the catapult(circle):
sin^2(t*theta)+cos^2(t*theta)=sqrt(r*sin(t*theta)+r*cos(t*theta))
Square both sides.
(sin^2(t*theta)+cos^2(t*theta))^2=r*sin(t*theta)+r*cos(t*theta)
sin^4(t*theta)+2sin^2(t*theta)*cos^2(t*theta)+cos^4(t*theta)=
r*sin(t*theta)+r*cos(t*theta)

The problem: how to find the derivative. If the derivative
could be found then it could be equated to the integral
of the second derivative.

Second derivative of position function:
s=100 rad/s^2



d=1000*t^2*cos(theta)
dd/dt=1000*( t^2*(dd/dt)*(-sin(theta))+2t*cos(theta) )
dd/dt=-1000*t^2*sin(theta)*(dd/dt)+2000t*cos(theta)
dd/dt(1+1000*t^2*sin(theta))=2000t*cos(theta)
dd/dt=(2000t*cos(theta))/(1+1000*t^2*sin(theta))
Maximum distance(critical number):
0=(2000t*cos(theta))/(1+1000*t^2*sin(theta))


Time to move on to the next problem. Good luck.

Answer 7

you would need to know the force exerted by the counterweight before you could even start the problem

Answer 8

i have a formula but didnt type it in. its written on my paper. you didnt have to be such an ***. go ahead and report me to yahoo. whats a few points anyways.