Motion in Two Dimensions. Projectile Motion A projectile is an object moving in two dimensions under...

Motion in Two Dimensions

Projectile Motion

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

ay = g

ax = 0



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.

If an object is launched at an initial angle of θ0 with the horizontal, the analysis is similar except that the initial velocity has a vertical component.

xv

yv

ovxv

yv1v

xv2v

xv

yv3v

xv

yv

fv

31 vv

fo vv



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

(A) is zero.

(B) remains constant.

(C) continuously increases.

(D) continuously decreases.

vo

x

y

atvv o

tgvvyoy

gtθsinvv oy Eq 2

Constant acceleration

θsinvv oyo

Eq 1 θcosovxv

Constant velocity



A ball is thrown with a velocity of 20 m/s at an angle of 60° above the horizontal. What is the horizontal component of its instantaneous velocity at the exact top of its trajectory?

(A) 10 m/s

(B) 17 m/s

(C) 20 m/s

(D) zero

vo

x

y

SubEq 1

t cosvx o Eq 3

1 Eq. cosvv ox

tΔxΔ

v

tx

vx

tx

θcosvo

Constant velocity


Practice Problem

If Vx = 6.80 units and Vy = 7.40 units, a) determine the magnitude of V.

b) determine the direction of V

2y

2x

2 vvv

2y

2x vvv 22 7.406.80

Vx

V

Vy

units 10

x

y

v

vtan

x

y1

v

vtan

80.640.7

tan 1 o47

vo

x

y

2

gtt θsinvy

2

o Eq 4

2at

tvxx2

oo

Constant acceleration

2

tgtv0y

2

yo

2

θsin2tg

tvy o

θsinvv oyo



A soccer ball is kicked with a velocity of 25 m/s at an angle of 45° above the horizontal. What is the vertical component of its acceleration as it travels along its trajectory?

(A) 9.80 m/s2 downward

(B) (9.80 m/s2) × sin (45°) downward

(C) (9.80 m/s2) × sin (45°) upward

(D) (9.80 m/s2) upward


When a football in a field goal attempt reaches its maximum height, how does its speed compare to its initial speed?

(A) It is zero.

(B) It is equal to its initial speed.

(C) It is greater than its initial speed.

(D) It is less than its initial speed.

voh

x

y

Sub into Eq 4

θcosvx

to

2

ooo θ cosv

xg2

1θsinθ cosv

xvy

θcosv2

gxθtan xy

22o

2 Eq 5

3 Eq. t θcosvx o

4 Eq. 2

gtt θsinvy

2

o

Solve Eq 3 for t


Vertical Position as a Function of Horizontal Displacement

voh

x

y

At the maximum

height (vy = 0)

0gtθsinvv oy

Sub into Eq 4

2g

θsinv y

22o

max Eq 7

4 Eq. 2

gtt θsinvy

2

o

2 Eq.gt θsinvv oy

g

θsinvt o

top Eq 6

2

o

oo

g

θsinv

2

g

g

θsinv θsinvy


Maximum Height

Problem

A football is kicked at ground level with a speed of 18.0 m/s at an angle of 35.0º to the horizontal. How much later does it hit the ground?

g

sinvt otop

Time in the air is twice the time to the top.

g

sinv2t 2t otopair

2m/s 8.9

35sinm/s 182 s 1.2

vo

R

hx

y

Eq.6

g

θsinvt otop

g

θsinv2t2t otopair

airxtvR

g

θsinv2θcosvR o

o

g

θ2sinvR

2o Eq 8

1 Eq. θcosvv ox


Range


At what angle should a water-gun be aimed in order for the water to land with the greatest horizontal range?

(A) 0°

(B) 30°

(C) 45°

(D) 60°

The range of a projectile is maximum (if there is no air resistance) for a launch angle of 45°.


A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5º above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile.

Problem

g2

sinv y

22i

max

2

22

m/s 8.92

5.34sinm/s 2.65

(b) the total time in the air

g

sinv2t 2t otopair

2m/s 8.9

5.34sinm/s 5.262 s 5.7

(c) the total horizontal distance covered (that is, the range).

g

θ2sinvR

2o

2

o2

m/s 8.9

5.342sinm/s 2.65 m 405

m 6.69

You are trying to hit a friend with a water balloon. He is sitting in the window of his dorm room directly across the street. You aim straight at him and shoot. Just when you shoot, he falls out of the window! Does the water balloon hit him?

a) yes, it hits

b) maybe—it depends on the speed of the shot

c) no, it misses

d) the shot is impossible

e) not really sure

Assume that the shot does have enough speed to reach the dorm

across the street.

Your friend falls under the influence of gravity, just like the water balloon. Thus, they are both undergoing free fall in the y-direction. Since the slingshot was accurately aimed at the right height, the water balloon will fall exactly as your friend does, and it will hit him!!

Question 3.10a Shoot the Monkey I

Shoot the Monkey

cosvv ox Eq 1Horizontal Velocity

gtsinvv oy Eq 2Vertical Velocity

t cosvx o Eq 3Horizontal Displacement

2

gtt sinvy

2

o Eq 4Vertical Displacement

Equations

22

o

2

cosv2

gxtan xy Eq 5

Vertical Position

g2

sinv hy

22o

Eq 7Maximum Height

g

2sinvR

2o

Eq 8Range

g

sinvt otop

Eq 6Time to

the Top

Equations

Motion in Two Dimensions. Projectile Motion A projectile is an object moving in two dimensions under...

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