RIVER CROSSING PROBLEMS PROJECTILE MOTION

SPH3U – Motion in 2 Dimensions

ADDING VELOCITIES IN 2-DIMENSIONS

Video – The River Boat Problem The keys to these types of problems:

Treat the x and y components independently The same time is taken for each motion !!!!!

PG. 74 – PRACTICE #2

BEIKiii'

GivenD= 20 . OnFx¥ 2.7 nls [w]Ty = 1 . 3 mls [ N ]

Required: *dtxAnalysis: Ty . 1}dy= jvxasdx

t

Steps=ddy_ ohiadx

Itat

at=y# Edxixat→20 .om[N] ?27÷[Dxk4

= m|s[N] [d×= 41.58mAst. 15.45 £dxs42m[w]

bt= loss

PROJECTILE MOTION

Projectile Motion the motion of a projectile under gravity

Projectile an object that moves along a two-dimensional curved trajectory in response to gravity

Time of Flight the time taken for a projectile to complete its motion

DESCRIBING PROJECTILE MOTION

2-dimensional vector problems velocity vectors pointing up or to the right have positive values velocity vectors pointing down or to the left have negative values

PROJECTILE MOTION – KEY POINTS

horizontal and vertical motions are independent the horizontal and vertical motions take the same amount of time projectiles move horizontally at a constant velocity projectiles undergo uniform acceleration in the vertical direction ( acceleration due to gravity)

PROJECTILE MOTION – KEY POINTS (CON’T)

objects can be projected horizontally or at an angle to the horizontal projectile motion can begin and end at the same or at different heights use the Big 5 equations of motion to solve projectile motion problems if you know the initial velocity and vertical displacement of the projectile you can solve for time of flight, maximum height, and range.

PG. 81 #8

*→

t

Givenszs

. Tdy = - kmday\Ej9.8÷[dadsur Y

sdx

Atiyeh's ,÷ln2D

Tins 45340525 )

-

RequiredAnalysis sty =D tx

Edy=hgstttaayat'

5¥ :.

hm=4s=¢nz5)attk(98⇒aE

s

- Rm : 1.90¥t -4.9mg at'

0 = - 4.9mg At 't 1.90 ;dtH2m

a = -4.9 b= 1.90 c=l2

Quadratic Formulaat . - b±Tb÷

Za

at . t.at#44.HzTtag

st = -1.90123M¥ao

at = -1.90+15.4 At= -1.90-15.4a a

st. -1 .↳s*wAoin$ 's t= ) . >6s✓

bty ' - Dtx - . 1.76 sTix = 4 . sz ¢os 25

° )tax . Ong OIdx ^

bdx = riot t §qµt2Idx =(4.sn#cos2sD(h6/DEdx=7.2m

✓

WORK FOR THE DAY

Pg. 78-81 – go over Sample Problem #1 and #2 Any issues with pg. 75 #8 and 9 ??? Pg. 81 #2, 4-7 Unit Test on Friday !! Quiz 6 this Wednesday

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