Projectile Motion:

is a form of motion in which an object or particle (called a projectile) is

thrown near the earth's surface, and it moves

along a curved path under the action of

gravity only. The only force of significance

that acts on the object is gravity, which acts

downward to cause a downward

acceleration. There are no horizontal forces

needed to maintain the horizontal motion.

This motion is simplified in to two independent motions:

Horizontal Motion in X-direction (ax=0)

Vertical Motion in Y-direction (ay=-g)

In projectile motion, gravity is the only force acting on the object

Projectile motion could be handled in the same way as two-

dimensional motion with constant acceleration analyzed:

Equation 1

rf=ri+vit +0.5at2 x y xf=xi+vxit yf=yi+vyit -0.5*g*t2

Since ax=0 ay=-g = -9.81 m/s2

Equation 2

rf=ri+0.5(vi+vf)t x y

xf=xi+0.5(vxi+vxf)t yf=yi+0.5(vyi+vyf)t

Equation 3

Vf2=Vi

2+2a(rf-ri) X y Vxf

2=Vxi2 Vyf

2=Vyi2-2g(yf-yi)

Since ax=0 ay=-g = -9.81 m/s2

Horizontal range

Xf - xi

= R = vxi t

yf – yi = v

yi t – ½ gt2 = 0 (this equation gives the required time for the

whole trip) Substitute the time into the first equation to evaluate the horizontal range.

Maximum Height

Note: the body reaches it's maximum height when Vy= ZERO; (a critical

point between decelerating when particle moves up and accelerating

when particle goes down)

Vyf=Vyi – gt = 0

t = Vyi/g (this equation gives the required time for particle to reach the

maximum height)

then substitute the time value into :

yf = yi + vyi – 0.5 gt2 => yf – yi = vyi – 0.5 gt2 = h

or substitute (vyf =0) directly in equation 3

h= vyi2/2g

Vya = 6.69