Motion in Two Dimensions:
Curved motion in a plane is motion in two dimensions. For example, if an object is to move along a semicircle as shown in the following figure, its position at any given time ( t ) can be determined by its two coordinates ( x ) and ( y ).
An excellent example of two-dimensional motion is the motion of a projectile. The motion of a projectile is a combination of two motions: (1) a constant speed motion in the horizontal direction, and (2) an accelerated motion in the vertical direction. For example, when a football is kicked or a cannonball fired, its shadow on a flat and horizontal ground moves at constant speed while its shadow on a vertical wall decelerates upward and comes to stop and then accelerates downward. Visualize a scenario in which the Sun is shining straight down and casts the shadow of the football on the ground while very bright horizontal light beams cast the shadow of the ball on a vertical wall as well.
The path of motion of the ball in air (neglecting air resistance) is a parabola. This parabolic motion can be replaced by a constant speed motion in the horizontal direction (x-axis) and an accelerated motion in the vertical direction (y-axis). The reason for the constant speed motion along the x-axis is that there is no gravitational pull horizontally to accelerate the ball and, neglecting air resistance, there is no force to slow it down. In the y-direction, because of the gravitational pull, it first slows down as it goes upward, and then speeds up as it returns to the ground. The equations of motion are:
|In the x-direction:||In the y-direction:|
x = (1/2) ax t2 +
Since ax = 0,
x = vix t.
|y = (1/2) g t2 + viy t,
g = (vfy -viy )/ t,
vfy2 - viy2 = 2 g y.
In the above equations, Vix and Viy are the horizontal and vertical components of the initial velocity vector. The velocity vector, itself, always acts tangent to the path of motion. Look at the following figure. It shows the path of a projectile that is thrown at velocity Vi = (Vi , θ) from Point O. As the object moves along the path, its x-component of velocity, Vix, remains constant, but its y-component of velocity, Vy, first keeps decreasing until the object reaches its highest elevation at C. Compare the lengths of the vectors used to show the magnitudes of Vy at points O, A, B, and C. At C, the highest point, Vy = 0. As the object passes Point C and starts descending, its y-component of velocity, Vy, keeps increasing in the downward direction until just before hitting the ground where it reaches is maximum value. If the ground is level and air resistance can be neglected, Vy at Point F will have the same magnitude as Viy at O, but opposite direction.
Answer the following questions: First think, then answer, and then click to check your answer.
1) As a projectile moves along its parabolic path, which velocity component does not change, Vx or Vy? Click here.
2) Does Vy increase or decrease as the object goes upward? Click here.
3) Is there any gravitational pull to accelerate the object in the x-direction? Click here.
4) If air resistance is negligible, is there any force to slow down the object in any direction? Click here.
5) Does the magnitude of Vy increase or decrease as the object goes downward from the highest point? Click here.
6) If the angle of throw ( θ ) and the initial velocity ( Vi ) are known, can Vix and Viy be calculated? Click here.
7) Write down the formulas that calculates Vix and Viy . Click here.
8) A cannon ball is fired at an initial velocity of Vi = (312m/s, 38.0o ). What are the components of the initial velocity in the x and y directions? (a) 312m/s, 312m/s (b) 246m/s,192m/s (c) 12m/s, 0.
You are standing in a 25m/s-moving train car holding a basketball. Let the closed car have a high enough ceiling for the ball to go as high as it needs. If you throw the basketball straight upward with respect to the train car at a vertical upward speed of 1.0m/s, determine if the following are (T) true or (F) false:
9) The ball will land in you hand after coming back down. (T) or (F)
10) The horizontal component of the ball's velocity is 25m/s. (T) or (F) Click here.
11) The initial vertical component of the ball's velocity is at 1.0m/s. (T) or (F)
12) An observer standing on the ground watching the process notices a parabolic path of motion for the ball not just a straight up-and-down motion as is observed by you. (T) or (F) Click here.
13) To the observer standing out of the train, the initial velocity vector is Vi = [25.02m/s , 2.3o] (T) or (F)
At this point, it is a good idea to start (objectively) from the beginning of this chapter, again.
Example 1:A cannon ball is fired at an initial speed of 250m/s through a 27o angle. Determine (a) its horizontal and vertical components of velocity, (b) the time it takes to reach the highest point, (c) the highest elevation it reaches, (d) the time it spends in air before landing, (e) the farthest horizontal distance it travels before landing, and (f) the equation of its path.
(a) Vix = 250cos(27o) = 223m/s; Viy = 250sin(27o) = 113m/s.
(b) At highest pt.: Vfy = 0; from g = (Vfy -Viy)/t ; t =(Vfy - Viy) /g = (0-113 )/-9.8 = 12s.
(c) At highest pt.: Vfy = 0; from (Vfy)2 - (Viy)2 = 2gy ; 02 -1132 = 2(-9.8)y ; y = 650m.
(d) Using the result of Part (b), ttotal = 2(12s) = 24s.
(e) Using the only x-equation, x = Vix t, x = ( 223 m/s)( 24s ) = 5400m.
(f) To find the equation of the path (a parabola), the common variable or the common parameter (t ), must be eliminated between the x- and y-components of motion. To do this, let us first write down the equations of motion in the x- and y- directions.
In the y-direction,y = (1/2) g t2 + Viyt. ; y = -4.9t2 + 113t.
In the x-direction,x = Vix t ; x = 223 t or t = x / 223.
Substituting fort in the y-equation, yields:
y = -4.9(x/223)2 + 113(x/223) or, y = -0.000099x2 + 0.51x. (A parabola)
Example 2: A small ball is rolling at a speed of 5.0m/s on a horizontal table 1.2m high with respect to the floor. How far from the edge of the table does it land after rolling off?
The components of initial velocity are:
Vix = [5.0m/s] cos0o = 5.0m/s; Viy = [5.0m/s] sin0o = 0.
If the origin is set at where the ball rolls off the table, and the positive y-axis is taken to be upward, then the ball lands at
y = -1.2m and g = -9.8m/s2.
It is easy to find the
falling time from:
since the ball is moving horizontally at the instant at rolls off,
Chapter 3 Test Yourself 1:
A football is kicked through a 30.0oangle at a speed of 15.0m/s. For answers Click here.
1) The x and y components of its initial velocity are: (a)7.50m/s, 13.0m/s (b)15.0m/s, 15.0m/s (c)13.0m/s, 7.50m/s.
2) The time it takes for the ball to reach its maximum height is (a)1.88s (b)1.33s (c) 0.765s.
3) On a flat and horizontal field, the time that the ball spends in air before landing is (a) 2.66s (b) 1.53s (c) 3.06s.
4) The maximum height it reaches is (a) 9.6m (b) 2.87m (c) 17.2m. Click here.
5) The distance it travels horizontally before landing is (a)19.9m (b) 9.95m (c) 4.9m.
A toy car pushed by a kid on a horizontal table 1.00m high with respect to the floor leaves the table's edge horizontally and strikes the floor, as shown. Answer the following questions.
Perform calculations before answering, if necessary.
6) The initial velocity vector is (a) Vi = (5m/s, 90o) (b) Vi = (5m/s, 45o) (c) Vi = (5m/s, 0.0o).
7) The time it takes for the car to travel the vertical height of 1.0m is (a)1/2 of the time it takes to travel the horizontal distance, x (b) equal to the time it takes to travel the horizontal distance, x (c) is twice the time for traveling the horizontal distance, x.
8) The initial speed in the y-direction is (a) 5m/s (b) 0 (c) 5tan(90o).
9) The falling time of the car is (a)1.0s ( b) 0.451s (c) 0.98s. (To check your answers Click here ).
10) The horizontal distance,x, that the car travels while in air is (a) 2.25m (b) 4.55m (c) 1.25m.
Problem:A kid is standing at a distance x from the edge of a building 37.0m high. He throws a rock through a 42oangle with the horizontal such that it almost passes at the top edge of the building horizontally. The kid's hand is 1.0m above the horizontal ground at the instant the rock leaves his hand. Let the kid be on the left side of the building and draw a rough sketch and show the half-parabola that is the rock's trajectory with the peak of the parabola being the top of the building. Set the x-y coordinates at the kid's hand and draw the initial velocity vector Vi with a throw angle of 42o. Show all distances and answer the following questions:
11) The y-component of the initial velocity is (a) Vi cos42o (b) Vi sin42o (c) Vi tan42o. Click here.
12) At the peak of the parabola, the y-component of velocityVy is (a) 0 (b) nonzero (c) (1/2)Vi .
13) In the y-direction, using: Vfy2 - Viy2 = 2gy results in (a) -Vi2 (sin42o)2 = -705.6 (b) -Vi2 (sin42o)2 = -725.2 (c) sinθ = 2.
14) Solving for Vi from question 13 results in (a) Vi = 39.7 m/s (b) Vi = 41.2 m/s (c) Vi = 12m/s. Click here.
15) The height of the building may be found using the fact that at the peak (a) Vix = 0 (b) Vy = 0 (c) Vy < 0.
16) The time it takes for the rock to reach the top is (a) 2.71s (b) 5.42s (c) 1.35s.
17)x, the kid's distance from the the foot of the building is (a) 40.0m (b) 20.0m (c) 80.0m. Click here.
Problem:A cannon ball is fired at an initial velocity of 320m/s through a 41oangle.
(18) Find the horizontal and vertical components of the initial velocity: Vix and Viy.
(19) Write the x- and y-components of equation of motion of the ball.
(20) Eliminate the common variablet between these two equations in order to arrive at the equation for its parabolic path. Click here.
Note:Before solving each problem, draw an appropriate figure for it.
1) A kid throws a tiny rock at the ground level at 14.0m/s through a 44.0oangle. Find (a) the horizontal and vertical components of the initial velocity, (b) the highest elevation it reaches, (c) the time the rock spends in air, and (d) the horizontal distance it travels.
2) Water from a fire hose, held at 60.0o with respect to the horizontal, comes out at a speed of 23.0m/s. The fireman adjusts his distance from the building to where water moves horizontally at the top edge of the building; in other words, the water jet forms exactly a half-parabola. Draw a figure for the problem and find (a) the height of the building and (b) the fireman's distance from the foot of the building knowing that the nozzle is held 1.0m above the ground. g = -9.81m/s2.
3) A tiny metal sphere rolling at 2.80m/s on a horizontal table rolls off the table's edge and falls on the floor at a point that is 1.40m from the foot of the table. Using g =-9.81m/s2, and neglecting air resistance, find (a) the falling time, and (b) the height of the table.
4) A cyclist is trying to jump over a number of busses parked side-by-side. The ramp angle is 32.0o and the maximum speed he can achieve as he leaves the take off ramp is 28.0m/s. The highest points on both the take-off ramp and the landing-ramp are at the same level as the busses' roofs are. If each bus including its spacing from the next bus requires a width of 3.00m, calculate the theoretical maximum number of busses that he can jump over. g=-9.81m/s2. Note that in reality (with air resistance present), the number of busses could be even less than half of what you calculate! For a tiny rock, air resistance may be neglected, but not for a cyclist.
5) Moving eastward, a car rolls off a vertical cliff and into a lake at an initial speed of 25.0m/s. The cliff angle is 10.0o below horizontal. The cliff edge is 6.60m above the lake surface. Find (a) its vertical speed just before hitting the lake surface, (b) the falling time, (c) the horizontal distance it travels before hitting the lake surface, and (d) the magnitude and direction of the velocity at which it enters the lake. g=-9.81m/s2.
1) (a) 10.1m/s, 9.73m/s (b) 4.82m (c) 1.98s (d) 20.0m
2) (a) 21.2m (b) 23.3m
3) (a) 0.500s (b) 1.23m
4) (a) 23
5) (a) 12.2m/s (b) 0.801s (c) 19.7m (d) 27.4m/s at -26.4o