(Work that shows the derivation of these
equations are shown below)
From these equations you can
easily find how far your projectile will go. Keep in mind that this is
in ideal conditions such as a particle in space. Drag on your
projectile will occur so your results will vary accordingly. In other
words the distance your projectile actually goes will be less than what you
predicted due to drag. Projectile drag calculations can get extensive
and are therefore left out in this example.
You fire a projectile from a
cannon being held 1.5 meters from the ground at an angle of 42°. Using
a chronograph you know that the velocity of the projectile leaving the
barrel was 115 meters/second. How long was the projectile in the air
for? How far will your projectile fire? What was the maximum height of the
projectile? What was the projectile's speed on impact?
t, x, ymax
and |v| at impact under the given conditions
cos(42) = 85.4617 m/s
v0y=115 sin (42) =
plug in values:
0=1.5+76.95t+(1/2)(-9.8)t2 solve for t.
t= 15.7236 seconds. So the projectile was in
the air for 15.7236 seconds
now use x=x0+v0xt
and plug in values:
x=0+85.4617 * 15.7236 = 1343.7656 meters. So the
projectile will go 1343.7656 meters if there was no drag or friction
the vertical velocity is zero when the projectile is at its highest point.
0=76.95-9.8*t solve for t, t=7.8520 seconds. Now put
that time into the y position equation:
plug in known value:
y=1.5+76.95t-(1/2)(9.8)t2 = 303.6073 meters.
So the projectile goes 303.6073 meters high
85.4617 m/s, vy = 76.95 - 9.8t = -77.1413 m/s (where
t=15.7236 seconds from above)
square root of (vx2 +
vy2) gives square root of (85.46172+
-77.14132)=36.8721 m/s So the projectile strikes the
ground at 36.8721 m/s.
Here is a nice little program that will give
you an idea for the motion of you projectile: