Projectiles Flashcards
horizontal plane Vx
ux
horizontal plane Sx
uxt
horizontal plane Vy
uy - gt
horizontal plane Sy
Uy.t - 1/2gt²
up an inclined plane Vx
ucosa - gtsina
up an inclined plane Sx
ucosa - 1/2gt²sinB
up an inclined plane Vy
usina - 1/2gtcosB
up an inclined plane Sy
utsina - 1/2gt² cosB
down an inclined plane Vx
ucosa + gtsinB
down an inclined plane Sx
utcosa + 1/2gt²sinB
down an inclined plane Vy
usina - gtcosB
down an inclined plane Sy
utsina - 1/2 gt²cosB
time of flight
sy=0
horizontal range
Sx @ sy=0
time to max height
Vy = 0
max height
Sy @ Vy = 0
direction of motion before max height
tan θ = Vy/Vx
direction of motion after max height
tan θ = -Vy / Vx
path contains point (a,b)
sx = a sy = b
up an inclined plane, what is H
the perpendicular height to the inclined plane
up an inclined plane, what is h
the perpendicular height the the original x-axis
how do you find h?
to find H; Sy @ Vy = 0
h = H/cosB
up an inclined plane, direction of motion before max height
tan θ = Vy/Vx
up an inclined plane, direction of motion after max height , Vx is positive
tan θ = -Vy/Vx
up an inclined plane, direction of motion after max height , Vx is negative
tan θ = Vy/Vx
strikes the plane at right angles, 2 things we know
Sy = 0 (time of flight) Vx = 0 (direction)
strikes the plane horizontally, 2 things we know
Sy = 0 (time of flight)
tan β̞ = -Vy/Vx (direction)
what is β̞
angle between Vx and horizontal
what is θ
angle between direction of motion (speed not velocity) and Vx
collisions parallel to the fixed surface
no change
collisions perpendicular to fixed surface
newtons experimental law
projected downwards Vx
ucosa
projected downwards Sx
ucosa t
projected downwards Vy
usin a + gt
projected downwards Sy
usina t + 1/2gt squared