http://www.ck12.org Chapter 3. Two-Dimensional Motion
FIGURE 3.27
The package remains under the plane,
but people on ground see the package’s
descent as a parabola. Author: Im-
age copyright Konstantin Yolshin, 2012;
modified by CK-12 Foundation - Christo-
pher Auyeung License: Used under
license form Shutterstock.com Source:
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Imagine a small airplane flying at a constant elevation of 80 m over your school soccer field, with a velocity of 50
m/s, (about to 110 mph), eastward; seeFigureabove. Ignore air resistance.
1.What would the motion of the package look like to an observer on the ground?
Answer: An observer on the ground would see the package fall in a parabolic arc, as if it had been projected
horizontally with a speed of 50 m/s; the same motion performed by the bullet and dart; (though the package would
remain directly below the airplane).
2.What would the motion of the package look like to an observer on the plane?
Answer:The package would appear to fall straight down, since the package and the airplane both have the horizontal
velocity of 50.0 m/s.
3.The plane releases the package when it flies over a target markedAon the ground, will the package land onA?
Answer:No, it will not, since, as the package falls, it is moving at the horizontal velocity 50 m/s.
4.How far from pointAdoes the package land?
Answer:
If we know how much time the package spends traveling at 50.0 m/s after it is released, we can usex=vtto solve
the problem.
The time the package remains in the air is found using our one-dimensional equation:
yf=^12 at^2 +viyt+yiwherea=g=−10 m/s^2 ,viy=0 m/s (at the instant the package is released it has no velocity
in they−direction),yf=0 andyi= 100 m; therefore,^12 (−10m/s^2 )t^2 + 0 +80m=0. Solving fort, we findt= 4 .0 s.
Thusx= ( 50 .0m/s)( 4 .0s) =200m. The package fell 200 m past targetA.
In order for the pilot to hit targetA, the package must be released 200m before reaching the target, or 4.0 seconds
before reaching targetA. SeeFigurebelow
Deriving some general results from the projectile equations
From the problems that we discussed, it should be reasonably clear that the equations we derived for one-dimensional
kinematics can be used, with caution, for 2-dimentional kinematics. The only difference is that care must be taken
in correctly identifying velocity components. Recall that if velocity is stated asVat an angle ofθ[U+F02C]then