2-dimentional motion Parabolic or Projectile Motion.

2-dimentional motion

Parabolic or Projectile Motion

Projectile Motion

A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

Motion in Two Dimensions

ay = g

ax = 0

Motion in Two Dimensions

Motion in Two Dimensions

Ignoring air resistance, the horizontal component of a projectile's acceleration

(A) is zero.

(B) remains a non-zero constant.

(C) continuously increases.

(D) continuously decreases.

Solving Problems Involving Projectile Motion

1. Read the problem carefully, and choose the object(s) you are going to analyze.

2. Draw a diagram.

3. Choose an origin and a coordinate system.

4. Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g.

5. Examine the x and y motions separately.

Solving Problems Involving Projectile Motion

6. List known and unknown quantities. Remember that vx never changes, and that vy = 0 at the highest point.

7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.

A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below. a) How high was the cliff?

Problem

y

d

v

b) How far from its base did the diver hit the water?

2at

tvx2

o

2gt

h2

2s 0.3m/s 8.9 22

m 44

vtx

tvd x s 0.3m/s 8.1 m 4.5

• A marble rolls off the edge of a table with a height of 0.755 m and strikes the floor at a distance of 24.3 cm from the edge of the table. Calculate the initial velocity of the marble.

• First convert cm to m, 24.3 cm = .243 m

- 0.755 m

0.243 m

g = - 9.81 m/s2

ax = 0dx = 0.243m (how far the ball traveled)

viy = 0g = - 9.80m/s2

dy = - 0.755 (a negative direction)

Find vix

• An archer stands on the wall of a castle and fires an arrow from a height of 12.10 m above the ground. If the archer fires an arrow parallel to the ground with an initial horizontal velocity of 11.0 m/s, how far will the arrow travel horizontally before hitting the ground?

- 12.10 m

g = - 9.81 m/s2

ax = 0vfx = 0vix = 11.0 m/s

viy = 0g = - 9.80m/s2

dy = - 12.10 (a negative direction)

Find dx

First find how long the arrow will be in the air

dy = 1/2gt2

t = 2dy = 2 (-12.10 m) = 1.57 s

-9.81 m/s2

Next, we can make use of the initial velocity in the horizontal and the change in time to find the displacement

vix = dx

tdx = vix t = (11.0 m/s)(1.57 s) = 17.3 m

An airplane is flying a practice bombing run by dropping bombs on an old shed. The plane is flying horizontally with a speed of 185 m/s. It releases a bomb when it is 593 m away from the shed, and it scores a direct hit. How high was the plane flying when it dropped the bomb?

593 m

ax = 0vfx = 0vix = 185 m/sdx = 593m

viy = 0g = - 9.81m/s2

Find dy

First we start by using the initial horizontal velocity and the horizontal displacement to determine how long the bomb was in the air

vix = dx dx = vix t t = dx 593m = 3.21 s t vix 185 m/s

Now that we know how long it takes the bomb to fall, we can calculate the bomb’s vertical displacement.

dy = 1/2gt2 = (0.5)(- 9.81m/s2)(3.21 s)2 = - 50.5 m(a negative direction)

A car launches horizontally off the edge of a cliff with a height of 18.6 m. It strikes the ground below at a distance of 98.4 m away from the edge of the cliff. How fast was the car going when it flew off the edge of the cliff?

98.4 m

ax = 0vfx = 0dx = 98.4 m

viy = 0g = - 9.8 m/s2

dy = - 18.6 m (a negative direction)

Find vix

- 18.6 m

A car launches horizontally off the edge of a cliff with a height of 18.6 m. It strikes the ground below at a distance of 98.4 m away from the edge of the cliff. How fast was the car going when it flew off the edge of the cliff?

98.4 m

ax = 0vfx = 0dx = 98.4 m

viy = 0g = - 9.8 m/s2

dy = - 18.6 m (a negative direction)

Find vix

- 18.6 m

First we start by using the vertical displacement and the acceleration due to gravity to determine how long the car was in the air

dy = 1/2gt2 t = 2dy = 2(-18.6 m) = 1.95s

g -9.81 m/s

Now that we know how long it takes the car to fall, we can calculate the car’s initial velocity.

Vix = 98.4 m = 50.5 m/s

1.95 s