These notes are provide detailed information on the projectile motion chapter in the Mind Action Series Physical Science Grade 12 textbook. These notes are colourful, have diagrams for further explanation and are very easily understood.
{ free fall } Chapter 4
projectile projected vertically
o x air resistance / other forces upwards which falls below og level:
therefore grav acc = 9.8 m.s-2
Vi @ A = upwards V decr as obj rises
@ max height, Vf = 0
Definition:
o If the object moves up and down max height when fall, Vi = 0
under the influence of the Vi at A = -Vi at C
gravitational force with no other
Vf = maximum (at ground/D)
force acting
o Object falls freely w/ Δt AB = Δt BC
gravitational acceleration
Δttotal = Δt AB + BC + CD
where g = 9.8 m.s-2 towards
the surface of the Earth Δy = distance of AD
a = k (9.8m.s-2)
{ going up & down }
Equations of motion to determine v, Δt & Δx
PROJECTILE EXAMPLES:
(Δy) bcz a = k (constant) in free fall
( l o o k @ d i a g r a m o n p g 3 1)
Vf = Vi + aΔt
TAKE DOWN = POSITIVE
Δy = ViΔt + ½aΔt2
Vi = 0 m.s-1 a = 9.8m.s-2 Δt =
Vf2 = Vi2 + 2aΔy 3s
➢ Choose direction (down / up) = positive & keep
this unchanged throughout. Δy = ?
➢ Indicate the direction chosen as positive at Δy = ViΔt + ½ a Δt2
the start of your answers.
Δy = (0)(3) + ½ (9.8)(3)2
Terminal velocity: Δy = 44.1 m (down)
o When force of air resistance
becomes = to weight of the obj, Vf = ?
\\
the Fnet is 0 & the obj will x longer Vf = Vi + aΔt
accelerate by will fall w/ constant v
Vf = 0 + (9.8)(3)
! NB ! Vf = 29.4 m.s-1 , down
➢ projectiles = same Δt to reach max height
from the point of upward launch to fall
back to the point of launch
Vi = 4.7 m.s-1 a = 9.8m.s-2 Δy = 44.1 m
EXPLANATION OF PROJECTILES Δt = ?
b
projectile projected vertically Vf2 = Vi2 + 2aΔy
upwards & falls back to same level: Vf2 = (4.7)2 + 2(9.8)(44.1)
Vi @ A = upwards direction V decr as obj Vf = 29.77 m.s-1 , down
rises
Vf = Vi + aΔt
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