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University Physics I - Classical Mechanics, 2019

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8.2. PROJECTILE MOTION 165<br />

The overall motion, then, is a combination of motion with constant velocity horizontally, and<br />

motion with constant acceleration vertically, and we can write down the corresponding equations<br />

of motion immediately:<br />

v x = v x,i<br />

v y = v y,i − gt<br />

x = x i + v x,i t<br />

y = y i + v y,i t − 1 2 gt2 (8.5)<br />

where (x i ,y i ) are the coordinates of the launching point (there is usually no reason to make x i<br />

anything other than zero, so we will do that below), and (v x,i ,v y,i ) the initial components of the<br />

velocity vector.<br />

By eliminating t in between the last two Eqs. (8.5), we get the equation of the trajectory in the<br />

x-y plane:<br />

y = y i + v y,i<br />

x −<br />

g x 2 (8.6)<br />

v x,i<br />

which, as indicated earlier, and as shown in Fig. 8.3, is indeed the equation of a parabola.<br />

The apex of the parabola (highest point in the trajectory) is at x max height = v x,i v y,i /g. We can get<br />

this result from calculus, or from a comparison of Eq. (8.6) with the canonical form of a parabola,<br />

or we can use some physics: the maximum height is reached, as usual, when the vertical velocity<br />

becomes momentarily zero, so solving the v y equation (8.5) fort max height and substituting in the x<br />

equation, we get<br />

2v 2 x,i<br />

t max height = v y,i<br />

g<br />

x max height = v x,iv y,i<br />

g<br />

y max height = y i + v2 y,i<br />

2g<br />

(8.7)<br />

The last of these equations should look familiar. It is, indeed a variation on our old friend vf 2 −v2 i =<br />

−2gΔy, only now instead of the full velocity ⃗v we have to use only the vertical velocity component<br />

v y . Just like for one-dimensional motion, this result follows again from conservation of energy:<br />

throughout the flight, we must have K + U G = constant, only now there is a component to the<br />

kinetic energy—the part associated with the horizontal motion—which remains constant on its<br />

own. In general, the kinetic energy of a particle will be 1 2 m|⃗v|2 ,where|⃗v| is the magnitude of the<br />

velocity vector—that is to say, the speed. In two dimensions, this gives<br />

K = 1 2 mv2 x + 1 2 mv2 y (8.8)

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