Ballista
Physics

Jason
Hoisington

Phys 212

3/21/05

Phys 212

3/21/05

The design of the
ballista is such that the force applied from the projectile comes from
the tension of the twisted ropes. The ropes, when the tension is
released, tend to return to their rest state with minimum tension, much
like a spring would expand after being compressed. Using this
similarity, the assumption can be made that the forces act in a similar
way, and that Hooke's Law can be applied to give at least a general
idea of the nature of the force applied by the ballista arms.
Doing so, we get

F=-kd,

^{2}=2ax,

where a is the acceleration and x is the length of the ballista that the projectile is accelerated upon. Since F=ma and F in this case is -kd, the equation can be simplified into

where F is the force applied, d is the
distance
that the arms are drawn back, and k is the force constant of the ropes
when under tension. Using this relationship assumes that the
force constant is constant, or that moving the arms back 2 meters gives
twice the force that moving them back 1 meter would do, which is most
likely not correct, but close enough for a general assumption of the
force to be made. When the force is applied, the projectile is
accelerated to the end of the ballista, at which point it released with
a velocity v and an angle q from the horizontal. The
velocity can be found using the kinematic equation

vwhere a is the acceleration and x is the length of the ballista that the projectile is accelerated upon. Since F=ma and F in this case is -kd, the equation can be simplified into

v^{2}=2-kdx/m.

After it is
released, the projectile obeys the laws of projectile motion,
disregarding air resistance. Therefore, the range of the ballista
can be given by the equation

R=v^{2}sin2q/g,

where g
is the
accleration due to gravity.