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)

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

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

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