1. HORIZONTALLY LAUNCHED PROJECTILE MOTION EQUATIONS VERTICAL MOTION
OBJECT FROM REST (Vi = 0)
Vfy = -gDt Velocity final in y-direction (given time)
Vfy = √-2gDy Velocity final in y-direction (given distance)
Dy = -1/2(gDt2) Distance in y-direction or height (given time)
ALSO
Dt = √(-2Dy)/g Time at given height
(some problems may use “h” in place of “Dy”)
HORIZONTAL MOTION
Vx = constant
Vx = Dx/ Dt
Dx = Vx Dt Distance in x-direction or range (given time)
solving for Dt in one dimension gives Dt in the other dimension
OBJECT WITH INITIAL VELOCITY (Vi ≠0)
Vfy = Viy -(gDt)
Vfy = √ Viy 2-(2gDy)
Dy = ViyDt -1/2(gDt2)

2. HORIZONTALLY LAUNCHED PROJECTILE MOTION EQUATIONS
MAGNITUDE OF RESULTANT VELOCITY CAN BE FOUND USING THE PYTHAGOREAN THEOREM

3. HORIZONTALLY LAUNCHED PROJECTILES
EXAMPLE PROBLEM: SOLVING FOR A HORIZONTALLY LAUNCHED OBJECT
P157 Glencoe, 2002
A stone is thrown horizontally at 15m/s from the top of a cliff 44m high.
How far from the base of the cliff does the stone hit the ground?
How fast is it moving the instant it hits the ground?
DIAGRAM
GIVEN UNKNOWN
EQUATION SOLVING FOR Dx SUBS/SOLVE Dx
SOLVE Dt SOLVE Vfy

4. HORIZONTALLY LAUNCHED PROJECTILES
EXAMPLE PROBLEM: SOLVING FOR A HORIZONTALLY LAUNCHED OBJECT
P157 Glencoe, 2002
A stone is thrown horizontally at 15m/s from the top of a cliff 44m high.
How far from the base of the cliff does the stone hit the ground?
How fast is it moving the instant it hits the ground?
GIVEN Vx and Vy SOLVE V