EXAMPLE PROBLEMS
Projectile Motion
A BALL IS THROWN HORIZONTALLY FROM THE
TOP OF A BUILDING 47.7 M HIGH. THE BALL
STRIKES THE GROUND AT A POINT 108 M FROM
THE BASE OF THE BUILDING.
A. Find the time the ball is in motion.
 B. Find the initial velocity
 C. Find the x component of the velocity just before it
hits the ground
 D. Find the y component of the velocity just before it
hits the ground

THE TIME THE BALL IS IN THE AIR
Is gravity effecting it?
 Yes, use the vertical equations.

D=1/2gt2
√2d/g
2*47.7/9.8
√9.7=3.12s
INITIAL VELOCITY OF THE BALL
V=d/t
108/3.12
V=34.6
FIND THE X COMPONENT OF THE VELOCITY
JUST BEFORE IT HITS THE GROUND

The same as the initial velocity
FIND THE Y COMPONENT OF THE VELOCITY
JUST BEFORE IT HITS THE GROUND
V=gt
V=-9.8*3.12
V=-30.5765
m/s
***Remember
it is falling down
and gravity would be negative
AN ARROW IS SHOT AT A 33⁰ ANGLE WITH THE
HORIZONTAL. IT HAS A VELOCITY OF 54 M/S
A. How high will it go?
 B. What horizontal distance will it travel?


54m/s

33⁰
vy
Vx
Sin 33 =y/54
Sin 33 *54
=29.41 m/s
Vertical speed
Cos 33=x/54
 Cos 33 * 54= 45.28m/s

Horizontal speed
Look at your vertical equations
 D=1/2gt2
 V=gt
 Let’s combine these two equations using substitution
method.
 t=v/g
 d=1/2 g(v/g)2
 d=44.13 m

NOW LETS SOLVE FOR THE HORIZONTAL
DISTANCE.
We solved for the horizontal velocity and now we
need to find an equation for the horizontal
distance.
 D=vt and because of the trajectory we need to use
twice the time from the previous problem.

A DESCENT VEHICLE LANDING ON THE
MOON HAS A VERTICAL VELOCITY VECTOR
TOWARD THE SURFACE OF THE MOON OF
31.6 M/S. AT THE SAME TIME, IT HAS A
HORIZONTAL VELOCITY OF
At
53.2 M/S.
what speed does the
vehicle move along its
descent path?
At what angle with the
vertical is its path?
AT WHAT SPEED DOES THE VEHICLE
MOVE ALONG ITS DESCENT PATH?


31.6 m/s
R
•53.6 m/s

a2+b2=c2
AT WHAT ANGLE WITH THE
VERTICAL IS ITS PATH?



31.6 m/s
R
θ=?
•53.6 m/s
tan θ=o/a
Θ=tan-1 (o/a)
JANET JUMPS OFF A HIGH
DIVING PLATFORM WITH A
HORIZONTAL VELOCITY OF 2.83
M/S AND LANDS IN THE WATER
2.2S LATER.
A.
How high is the platform?
B. How far from the base of
the platform does she land?
A IS ASKING FOR THE VERTICAL DISTANCE
2
D=1/2gt
B IS ASKING FOR HORIZONTAL DISTANCE

D=vt