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A projectile of mass 0.552 kg is shot from a cannon, at height 7 m,?
A projectile of mass 0.552 kg is shot from a
cannon, at height 7 m,
with an initial velocity vi having a horizontal
component of 5 ms.
The projectile rises to a maximum height of
Y above the end of the cannon's barrel and
strikes the ground a horizontal distance X
past the end of the cannon's barrel.
The acceleration of gravity is 9.8 ms^2 :
Determine the vertical component of the
initial velocity at the end of the cannon's bar-
rel, where the projectile begins its trajectory.
Answer in units of ms.
Determine the maximum height Y the pro-
jectile achieves after leaving the end of the
cannon's barrel. Answer in units of m.
Find the magnitude of the velocity vector
when the projectile hits the ground. Answer
in units of ms.
Find the magnitude of the angle (with repect
to horizontal) the projectile makes when im-
pacting the ground. Answer in units of degree.
Find the range"x" of the projectile. Answer
in units of m.
The equations for this are:
y(t)=7+v0y*t-.5*g*t^2
and
x(t)=5*t
Step one is to determine the total flight time.
The time to apogee, where y(t1)=y, is
0=v0y-g*t1
simplify
v0y/g=t1
the time from apogee to ground,t2, is found by
y+7=.5*g*t2^2
t2=sqrt(2*(y+7)/g)
the total flight time is t1+t2, so
x(t)=5*(v0y/g+sqrt(2*(y+7)/g))
The apogee, y, (above the cannon barrel) is
y=v0y*t1-.5*g*t1^2
since v0y/g=t1
y=v0y^2/g-.5*v0y^2/g
or
y=v0y^2/(2*g)
ad 7 m to get the height above ground
The angle at impact is found by looking at vy and vx at impact
and taking atan(vy/vx)
since vx is constant at 5
and vy=g*t2
or
vy=g*sqrt(2*(y+7)/g)
or
sqrt(2*g*(y+7))
to get actual numbers, we need either to know y, or the angle of the cannon barrel, or the magnitude of velocity of the projectile when leaving the cannon.
j
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