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