3.4. Projectile Motion http://www.ck12.org
FIGURE 3.23
Author: Soccer player: Image copy-
right Mandy Godbehear, 2012; Dia-
gram: CK-12 Foundation - Christo-
pher Auyeung License: Used under
license from Shutterstock.com Source:
http://www.shutterstock.comFinding instantaneous velocity
Let’s look back at the case of a dart fired out from a dart gun, using vector mechanics. Consider the trajectory of
the dart sometime between pulling the trigger and striking the ground. The diagram below is a representation of the
instantaneous velocity components of the dart.
FIGURE 3.24
Author: CK-12 Foundation - Raymond
Chou License: CC-BY-NC-SA 3.0Using only one-dimensional mechanics, we found that in the y-direction, it takes 0.55 seconds for the dart to hit
the floor from a height of 1.5 meters. If the dart traveled a horizontal distance of 6.0 m in 0.55 seconds, then its
horizontal component of velocity is therefore
vx=6 .0m
0 .55s
= 10 .9 m/s.
What is thevector velocityat time,t= 0 .25s?
Thexandyvelocity components represent the legs of the right triangle (seeFigureabovevx^2 +vy^2 =v^2 , gives the
magnitude of the instantaneous velocity of the projectile once the square root is taken.
Thexvelocity is constant, so we know it is 10.9 m/s at all times. For theyvelocity, we know from one-dimensional
motion thatvf=at+vi. The acceleration from gravity is constant, giving:(−10m/s^2 )( 0 .25s)+0m/s=− 2 .5 m/s.
Now that both legs of the right triangle are known, we can apply the Pythagorean Theorem to solve for the
instantaneous speed (the hypotenuse of the right triangle) at time,t= 0. 25 s.
v=√
v^2 x+v^2 y=√
( 10 .9m/s)^2 +( 2 .5m/s)^2 = 11 .2 m/s