Topic 4: Projectile Motion



Topic 4: Projectile Motion

An object that moves through the air without a propulsion system is referred to as a projectile. The curved path followed by the projectile is called the trajectory and forms a parabola. The horizontal displacement of a projectile is called the horizontal range ((dx).

In projectile motion, the projectile travels at a constant velocity in the horizontal direction only. In the vertical direction, all projectiles accelerate downward at 9.8 m/s2 due to the force of gravity. A projectile experiences both a uniform horizontal motion (constant horizontal velocity) combined with a vertical acceleration (assuming no air resistance or other forces).

Since the horizontal and vertical motions are independent of each other, the motion of the projectile can be analyzed by considering its vertical and horizontal components. The vertical component is the same as simple falling body motion and the horizontal motion is not affected by gravity. The curved path (parabola) is the result of the two motions occurring simultaneously.

Equations for Projectile Motion

Horizontal Motion

[pic] or [pic] (since ax = 0)

Vertical Motion

[pic] or [pic]

[pic]

[pic]

[pic]

[pic]

Assignment

unit 1 practice 2, # 3, 5

unit 1 practice 2

Problem: A projectile launched Horizontally

A rock is thrown horizontally at 10.0 m/s from the top of a cliff 122.5 m high.

a) time of flight (the time taken to reach the ground)

b) range (the horizontal distance travelled by the ball)

c) final velocity (just before it strikes the ground)

vix = 10.0 m/s

viy = 0

ax = 0

ay = -9.8 m/s2

(dy = -122.5 m

a) [pic]

-122.5 m = 0 + ½ (-9.8 m/s2)((t)2

(t = 5.0 s

(When taking the square root, take the positive value since time must be positive.)

The ball flight is 5.0 s.

b) [pic]

= 10.0 m/s x 5.0 s

= 50 m

The range is 50 m.

c) For horizontal motion there is no acceleration, therefore, the initial and final velocities are equal (assuming no friction). vix = vfx = 10.0 m/s

For vertical motion, the object is accelerating.

[pic]

Since the ball rolled off the cliff, its initial motion is horizontal, therefore, the vertical initial velocity (viy) is zero.

[pic]= 0 + 2 x (-9.8 m/s2) (-122.5 m)

vfy = 49.0 m/s

[pic]

= (10.0 m/s)2 + (49.0 m/s)2

vf = 50 m/s

tan ( = vfy / vfx

= (49.0 m/s) / (10.0 m/s)

( = 780

The final velocity is 50 m/s [to the right and 780 to the horizontal]

Problem: A projectile Launched at an Angle (and ∆dy = 0)

A golfer strikes a golf ball on level ground. The ball leaves the ground with an initial velocity of 42 m/s at 320 above the horizontal. Determine

a) ball’s time of flight (the time taken to reach the ground)

b) horizontal displacement (range).

c) maximum height

vi = 42 m/s [320 above the horizontal]

θ = 320

ax = 0

ay = -9.8 m/s2

(dy = 0

a) Vix = vi cos 320 Viy = vi sin θ

= 42 m/s x cos 320 = 42 m/s x sin 320

= 35.6 m/s = 22.3 m/s

For vertical components

[pic]

0 = 22.3 m/s ∆t + ½ (-9.8 m/s2) (∆t)2

0 = 22.3 m/s + ½ (-9.8 m/s2) (∆t) (divide both sides by ∆t)

∆t = 2 (-22.3)/ -9.81

= 4.55 s

The ball was in flight for 4.6 s

b) For the horizontal displacement, there is no acceleration

ax = 0

∆dx = Vix ∆t

= 35.6 m/s x 4.55 s

= 161.98 m

The horizontal displacement (range) is 162 m.

c) For maximum height

[pic]

0 = (22.3)2 + 2(-9.8)(∆dy)

∆dy = 25.4 m

The maximum height is 25 m.

Problem: A projectile Launched at an Angle (and ∆dy ≠ 0)

An astronaut on the Moon, where |g| = 1.6 m/s2, strikes a golf ball giving it a velocity of 32 m/s [350 above the Moon’s horizontal]. The ball lands in a crater floor that is 15 m below the level where it was struck. Determine

a) ball’s time of flight (the time taken to reach the ground)

b) range (the horizontal distance travelled by the ball)

vi = 32 m/s

θ = 350

ax = 0

ay = -1.6 m/s2

(dy = -15 m

a) vix = vi cos θ viy = vi sin θ

= 32 m/s x cos 350 = 32 m/s x sin 350

= 26.2 m/s = 18.4 m/s

For vertical components

[pic]

-15 m = 18.4 m/s ∆t + ½ (-1.6 m/s2) (∆t)2

(-0.8 m/s2)(∆t)2 + 18.4 m/s ∆t +15 = 0 (rearrange - quadratic equation form)

since [pic] (quadratic equation solution)

[pic]

∆t = (-18.4±19.7) ÷ (-1.6)

∆t = 23.8 s (taking only the positive value)

The ball was in flight for 24 s.

b) For the horizontal displacement, there is no acceleration

ax = 0

∆dx =vix ∆t

= 26.2 m/s x 23.8 s

= 624 m

The horizontal displacement is 620 m.

Assignment

unit 1 practice 3, # 9,10, 3

unit 1 practice 3

p.85 # 3

3) A fighter jet going 350 km/h dives at an angle of 250 to the vertical. If it drops a bomb from a height of 200 m, calculate:

a) how far the bomb will travel before

hitting the ground (range).

b) the velocity of the bomb just before it hits the ground.

answers

3a) 84 m

3b) 116 m/s [690 to horizontal]

-----------------------

could use an air table at an angle

consider from Pasco, shoot the target 22 (me-6826) $270

p46 Nelson

vi

(dy = -122.5 m

(

range

down is negative

(dx

vf

(

vf

vfy

vfx

vi

(

range

down is negative

ÆDÈDÊDÌDEE-E E4E6E@EBEÖEØEFFF

FbFdF’F”F–F˜FHåÒÁ´¦´¦´¦´¦´Á´ŒyÁ´Á´aNÁ´%jp[?]CJEHàÿU[pic]^J[?]aJ/j¾³G[pic]h9,h>[?]CJOJQJU[pic]V[pic]nH tH Only when ∆dy = 0

the range can also be calculated by using

[pic]

= [(42 m/s)2 (sin 2 x 320)]/(9.81 m/s2)

= 162 m

(Note that at the highest point, [pic] m/s. This also happens when an object thrown directly up reaches the top of its flight.)

(

vi

(dy = -15 m

range

(dx

viy

vix

down is negative

p50

Nelson

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download