Motion in Two Dimensions
It is essentially impossible to find an object moving in a straight path. Motion in at least two, usually three, dimensions is the norm. We will limit ourselves to two dimensional motion.
Horizontal and Vertical Motions The conceptual barrier to understanding two dimensional (arc) type of motion is realizing that it can be separated into the vertical and horizontal components and that they can be investigated separately. How fast an object ascends or descends is unrelated to its horizontal speed. In a classic demonstration of this, hold two marbles the same distance above the floor. As one is just dropped, flick the other forward. They strike the floor simultaneously! Of course, the second one lands some distance from the launch point but they both fell the same vertical distance in the same time. Your finger energy affected the horizontal motion and gravity controlled the vertical motion. Because gravity affected both marbles equally, they accelerated downward at the same rate, moving faster and faster until they impacted the floor simultaneously. Did the horizontal speed change? No – your finger was like a miniature catapult. Once the marble was beyond the end of your finger, its speed was set. While watching the flight of a golf ball or football, our eyes can not separate the vertical and horizontal motions but slow motion can.
Here’s a neat thought demonstration to help understand this idea of vertical and horizontal motions combining to create motion in the shape of an arc. Imagine: inside a huge white warehouse, a robotic arm holding a golf club, a red golf ball teed up, a robot car out in front of the tee. The little car has a speedometer and a camera and radar pointing straight up. The arm swings, the golf ball arcs away and the car takes off at just the right speed to keep it under the ball during its flight. What does the camera see? A red ball getting smaller and smaller as it climbs and then larger and larger as it descends. It looks like the ball is just moving straight up and down – we can not even tell it is moving forward as it climbs and falls. What does the radar tell us? As the ball rose higher and higher, its speed dropped off until, for an instant it climbed no higher. It seemed momentarily suspended in midair. Then, the ball began to descend, hardly moving at first but gaining speed until, just before it hit the car, it was moving as rapidly as when it first started to climb. Why the continual speed change as the ball ascends and descends? Gravity; it is always tugging down on things, trying to make them move more and more slowly as they ascend and, faster and faster as they descend. What does the speedometer read during the car’s trip along the floor of the warehouse? It was constant because gravity has no effect on the horizontal component of the ball’s motion. Just as with the marble demonstration above, once the ball leaves the club face, its speed is set and does not change.
This unusual demonstration summarizes the following key things about objects moving in arcs:
i) distance up = distance down
ii) initial speed up = final speed down
iii) final speed up = initial speed down = 0 m/s
iv) time up = time down
v) horizontal speed is constant.
Here’s another, more concrete demo about arc motion. Get a piece of looseleaf and a marker. Practice making smooth up and down motions and then, place the marker at a spot near the edge of the page. Pull the page slowly across in front of you and, at the same time, move the marker smoothly up and back down. What shape do your horizontal and vertical motions combine to create? An Arc.
Problem Types Here is a selection of standard arc motion problems. We use equations from the unit on Acceleration. Work first with the vertical and then with the horizontal motions. Reread the five points above to see when velocities are equivalent or when they are zero. Note: at the top of the arc, for a fraction of a second, an object’s final speed up is 0 m/s because it has finished climbing and, since it has not yet begun to fall, its initial speed down is 0 m/s. Also, recall that a deceleration is negative and acceleration is positive. So, when an object is ascending, a is – because gravity is slowing the object. When an object is descending, a is + because gravity is increasing their speed. For objects moving in arc’s we must find the vertical (Vv) and horizontal (Vh) components of the launch velocity.
1. Straight up and down. A ball is tossed up at 12 m/s.
t up: Vf = Vi + at
(Vf – Vi)/a = t
(0 m/s – 12 m/s) / – 9.8 m/s2 = t = 1.22 s
d up: d = Vi t + ½ at2
12 m/s * 1.22 s + ½ ( – 9.8 m/s2) 1.222 s = 7.35 m
t total: t up = t down therefore t total = 1.22 s * 2 = 2.44s
2. a) Arc. A soccer ball is kicked at 7 m/s, 250. Use Vv for vertical motion and Vh for horizontal motion.
Vv = sin 250 * 7 m/s = 2.96 m/s Vh = cos 250 * 7 m/s = 6.34 m/s
t up: Vf = Vi + at
(Vf – Vi)/a = t
(0 m/s – 2.96 m/s) / – 9.8 m/s2 = t = 0.3 s
d up: d = Vi t + ½ at2
2.96 m/s * 0.3 s + ½ ( – 9.8 m/s2) 0.3 2 s = 0.45 m
t total: t up = t down therefore t total = 0.3 s * 2 = 0.6 s
d downrange: dh = Vh t = 6.34 m/s * 0.6 s = 3.8 m
b) Please redo this question but change the launch angle to 650. Compare the results with 2. a).
The results from 2. b) point out something counter-intuitive: the soccer ball will travel the same distance downrange if it were launched at 650. How can this be? Should the projectile not plop into the ground much closer to the launch point if it is takes off up into the air at a steeper angle? It works this way: the steeper launch angle of 650 creates a larger vertical velocity component and a smaller horizontal velocity component than when it is launched at 250. The large vertical velocity component boosts the projectile quite high into the air so its ascent and descent travel times are large. And, because the projectile spends more time in the air, the smaller horizontal velocity component is able to move it the same distance downrange. See that the distance downrange depends on both the total time in the air and the horizontal velocity component. Think of it this way: one person walks faster for a short time period and another person walks more slowly but for a longer time period — they can both cover an equal distance. So, for any launch angle less than 450, subtract it from 900 to find the complimentary launch angle that will create the same downrange distance.
3. Arc toward a barrier. A catapult launches a rock at 17 m/s, 400 toward a castle wall 6 m away. How
high up a castle wall does the rock impact?
Vv = sin 400 * 17 m/s = 10.93 m/s Vh = cos 400 * 17 m/s = 13.02 m/s
t barrier: dh = Vh t
dh / Vh = t
6 m / 13.02 m/s = t = 0.46 s
d up: d = Vi t + ½ a t2
10.93 m/s * 0.46 s + ½ (– 9.8 m/s2) 0.46 2 s = 3.99 m
4. Falling or dropped from a height. A pebble falls from a 50 m cliff.
t down: d = Vi t + ½ at2
50 m = 0 m/s * t + ½ 9.8 m/s2 t2
50 m / 4.9 m/s2 = t2 = 10. 2 s2 therefore t = 3.19 s
Vf: Vf = Vi + at = 0 m/s + 9.8 m/s2 * 3.19 s = 31.26 m/s
5. Being thrown down from a height. A tourist on the edge of a 20 m cliff throws down a pebble at 3 m/s.
t down: d = Vi t + ½ at2
20 m = 3 m/s t + ½ 9.8 m/s2 t2
20 m = 3 m/s t + 4.9 m/s2 t2
0 = 4.9 m/s2 t2 + 3 m/s t – 20 m (Use quadratic solution method: t = – b + √b2 – 4ac )
2a
t = – 3 + √3 2 – 4 * 4.9 * –20 = 1.74 s (the other answer, – 23 s, is undefined)
2 * 4.9 m/s2
Vf: Vf = Vi + at = 3 m/s + 9.8 m/s2 * 1.74 s = 20.05 m/s
6. Arc off a height. An object is launched at 8 m/s, 350 from the edge of a 40 m cliff.
Vv = sin 350 * 8 m/s = 4.59 m/s Vh = cos 350 * 8 m/s = 6.55 m/s
t up: Vf = Vi + at
(Vf – Vi)/a = t
(0 m/s – 4.59 m/s) / – 9.8 m/s2 = t = 0.47 s
d up: d = Vi t + ½ at2
4.59 m/s * 0.47 s + ½ ( – 9.8 m/s2) 0.47 2 s = 1.07 m
d total: d total = d up + cliff = 1.07 m + 40 m = 41.07 m
t down: d = Vi t + ½ at2
41.07 m = 0 m/s * t + ½ 9.8 m/s2 t2
41.07 m / 4.9 m/s2 = t2 = 8.38 s2 therefore t = 2.9 s
t total: t up + t down therefore t total = 0.47 s + 2.9 s = 3.37 s
d downrange: dh = Vh t = 6.55 m/s * 3.37 s = 22.07 m
7. Arc across a gulf. An object is launched at 9 m/s, 300 from the edge of a 60 m cliff. Will it land on a
4 m tall surface 20 m away?
Vv = sin 300 * 9 m/s = 4.5 m/s Vh = cos 300 * 9 m/s = 7.79 m/s
t up: Vf = Vi + at
(Vf – Vi)/a = t
(0 m/s – 4.5 m/s) / – 9.8 m/s2 = t = 0.46 s
d up: d = Vi t + ½ at2
4.5 m/s * 0.46 s + ½ ( – 9.8 m/s2) 0.46 2 s = 1.03 m
d total: d total = d up + cliff – surface = 1.03 m + 60 m – 4 m = 57.03 m (pretend the 4 m high surface
butts up against the cliff foot and see if the object has covered the 20 m when it is level with the
surface)
t down: d = Vi t + ½ at2
57.03 m = 0 m/s * t + ½ 9.8 m/s2 t2
57.03 m / 4.9 m/s2 = t2 = 11.64 s2 therefore t = 3.41 s
t total: t up + t down = t total = 0.46 s + 3.41 s = 3.87 s
d downrange: dh = Vh t = 7.79 m/s * 3.87 s = 30.15 m (See that the object has crossed the 20 m gap
with an extra 10.15 m by the time it has fallen to the height of the surface.)
8. Shoot straight out from a height. A rock is tossed straight out at 6.7 m/s from the edge of a 35 m cliff.
t down: d = Vi t + ½ at2
35 m = 0 m/s t + ½ 9.8 m/s2 t2
35 m / 4.9 m/s2 = t2 = 7.14 s2 therefore t = 2.67 s
dh: dh = Vh t
6.7 m/s * 2.67 s = 17.89 m