The document discusses projectile motion and provides sample problems to illustrate the concepts. It begins by defining projectile motion and describing the forces acting on a projectile. It then presents the kinematic equations for the horizontal and vertical motion of a projectile. Several sample problems are worked out applying these equations to calculate values like minimum launch speed, projectile impact location and time, and velocity at impact.

1. Dynamics of Rigid Bodies

2. DISCUSSION
 Projectile motion is one of the traditional
branches of classical mechanics, with
applications to ballistics. A projectile is any body
that is given an initial velocity and then follows a
path determined by the effect of the gravitational
acceleration and by air resistance. Projectile
motion is the motion of such a projectile.
 The free-flight motion of a projectile is often
studied in terms of its rectangular components.
To illustrate the kinematic analysis, consider a
projectile launched at point (Xo, Yo), with an
initial velocity of Vo, having components (Vo)x
and (Vo)y.

3.  In the figure, when air resistance is neglected, the
only force acting on the projectile is its
weight, which causes the projectile to have a
constant downward direction of approximately a =
g = 9.81m/s² or g = 32.2ft/s².

4.  HORIZONTAL MOTION. Since ax =
0, application of the constant acceleration
equations, yields
 +→ v = vo + act ; vx = (vo)x
 +→ x = xo + vot + ½ act² ; x = xo + (vo)xt
 +→ v² = vo² + 2ac (x-xo) ; vx = (vo)x
 The first and last equations indicate that the
horizontal component of velocity always remains
constant during the motion.

5.  VERTICAL MOTION. Since the positive y-axis is
directed upward, then ay = -g. Applying the
equations, we get
 +↑ v = vo + act ; vy = (vo)y - gt
 +↑ y = yo + vot + ½ act² ; y = yo + (vo)yt + ½ gt²
 +↑ v² = vo² + 2ac (y-yo) ; vy² = (vo)y² - 2g(y-yo)
 Recall that the last equation can be formulated on the
basis of eliminating the time t from the first two
equations, and therefore only two of the above three
equations are independent of one another.

6. SAMPLE PROBLEMS
Problem No. 1
Problem No. 2
Problem No. 3
Problem No. 4
Problem No. 5

7. Problem No. 1
 A daredevil tries to jump a canyon of width 10 m.
To do so, he drives his motorcycle up an incline
sloped at an angle of 15 degrees. What
minimum speed is necessary to clear the
canyon?
 (Sample Problems) Solution>>>

8. Solution to Problem No. 1
 First, write the constant velocity equation for the horizontal direction:
 vx = Dx / Dt = V cos15o
 Where V represents the total velocity at the 15o angle.
 Solving for V, V = Dx / (Dt cos15o ) Equation 1
 Now write the constant acceleration equation for the vertical
direction:
 Dy = (1/2) a Dt2 + vy Dt where vy = V sin15o
 substituting for vy:
 Dy = (1/2) a Dt2 + V sin15o Dt Equation 2
 Now combine equations 1 and 2 by substituting for V:
 Dy = (1/2) a Dt2 + Dx / (Dt cos15 ) sin15o Dt
 simplify
 Dy = (1/2) a Dt2 + Dx sin15o / cos15o

9.  Solve for Dt:
 Dt = [ 2( Dy - Dx sin15o / cos15o ) / a ]1/2
 Substitute values:
 Dt = [ 2( 0m - 10m sin15o / cos15o ) / 10 m/s2 ]1/2
 Dt = 0.73s
 Now insert the time into equation 1 to find V.
 V = 10m / (.73s) * cos15o)
 V = 14.18 m/s, the minimum speed needed for success.
 (Sample Problems)

10. Problem No. 2
 An example of a projectile
motion is shown in the
figure, which shows a car
rolling off a cliff. Suppose
the care leaves the cliff
with a velocity of 10m/s
directed along the
horizontal. If the cliff has a
height h=20 m where and
when will the car land?
 (Sample Problems)
 Solution>>>

11. Solution to Problem No. 2
 y = h - ½gt2
 0 = 20m - ½ (9.81m/s2)t2
 t = 2.02secs
 x = Voxt
 x = 10m/s(2.02secs)
 x = 20.20meters
(Sample Problems)

12. Problem No. 3
 A projectile shot at an angle of 60o above the
horizontal strikes a building 80 ft away at a point
48 ft above the point of projection. (a) Find the
initial velocity, (b) Find the magnitude & direction
of the velocity when it strikes the building.
 (Sample Problems) Solution>>>

13. Solution to Problem No. 3
 The wording identifies the problem as a projectile motion problem. We
draw a figure, choose a CS, and write down the initial conditions &
acceleration in the problem.
 As before, since this is a 2-dimensional problem, initial position, initial
velocity, and acceleration are specified by two numbers:
 xo = 0 vox = vo cos 60 = (.5) vo ax = 0 .
 yo = 0 voy = vo sin 60 = (.866) vo ay = - 32 ft/s2 .
 We note that the quantity vo is not given in the problem. Hence, our first
task will be to determine this quantity. Inserting these values into the
general equations of motion in 2-dimensions, we have:
 x(t) = (.5) vo t y(t) = - (1/2)(32) t2 + (.866)vo t
 vy(t) = - 32 t + (.866)vo .
 Since vo is not given in the problem, some
other piece of information must be given.
We read that the projectile: "strikes a building
80 ft away at a point 48 ft above the point of
projection". Drawing a figure, we let the
instant when the projectile strikes the building
be: t'. Then we have:

14.  x(t') = 80 = (.5) vo t‘
 y(t') = 48 = -(1/2)(32)t'2 + (.866)vo t' .
 Hence we have 2 equations in 2 unknowns and can solve for both t' and
vo. Solving we find: t' = 2.38 seconds; vo = 67.3 ft/sec.
 Since we now know vo, then our specific equations of motion are
complete, and we can calculate any other quantity associate with the
motion. We are specifically asked for the velocity when it strikes the
building (at time t = 2.38 sec.). Thus:
 vx = vox = (.5)(67.3) = 33.6 ft/sec
 vy(t=2.38s) = - 32(2.38) + (.866)(67.3) = - 17.9 ft/sec .
 We then draw the velocity vector from its components calculated above.
The magnitude & direction (angle) can then be determined:
 v=
= 38.1 ft/sec.
 tan = (17.9)/(33.6)  = 28o
 (below hor. as shown)
 (Sample Problems)

15. Problem No. 4
 A bullet has a speed of 350 m/sec as it leaves a
rifle. If it is fired horizontally from a cliff 6.4 m
above a lake, how far does the bullet travel before
striking the water?
(Sample Problems) Solution>>>

16. Solution to Problem No. 4
 We have a 2-dimensional problem with constant acceleration
(acceleration due to gravity). This is a projectile motion problem. The
figure is as shown and the coordinate system selected is drawn. The
origin is placed at the bullet's location at time t=0. Hence the initial
conditions for the problem are:
 x(t=0) = xo = 0 ; y(t=0) = yo = 0
 vx(t=0) = vox = 350m/s ; vy(t=0) = voy = 0
 Since the only force acting is gravity (downward = + y direction), we
have: ax = 0; ay = + g = + 9.8 m/sec2. The general solutions for the
constant acceleration problem in two dimensions are:
 x(t) = (1/2) ax t2 + vox t + xo y(t) = (1/2) ay t2 + voy t + yo
 vx(t) = ax t + vox vy(t) = ay t + voy

17.  Inserting the values of acceleration and the initial conditions gives
us the specific equations (applicable to this one particular problem).
 x(t) = (350) t y(t) = (1/2)(9.8) t2
 vx = 350 m/s vy(t) = 9.8 t
 Let t' be the time when the bullet hits the lake. We then know that:
y(t') = + 6.4 m. Thus:
 y(t') = + 6.4 = + 4.9 t'2  t' = 1.143 sec.
 The horizontal (x) position of the bullet at this time is then:
 x(t') = (350)(1.143) = 400 m.
(Sample Problem)

18. Problem No. 5
 A player kicks a football at an angle of 37o with the
horizontal and with an initial speed of 48 ft/sec. A
second player standing at a distance of 100 ft from
the first in the direction of the kick starts running to
meet the ball at the instant it is kicked. How fast
must he run in order to catch the ball before it hits
the ground?
 (Sample Problems) Solution>>>

19. Solution to Problem No. 5
 We have a projectile motion problem (as far as the football is concerned).
Hence we have drawn a figure, chosen a CS, and write down the initial
conditions (initial position & velocity) of the football (at t=0).
 x0 = 0; y0 = 0; v0x = v0 cos 37; v0y = v0 sin 37
 The acceleration is: ax = 0; ay = - 32 ft/sec2.
 The general equations of motion for constant acceleration in 2-dimensions
are:
 x(t) = (1/2) ax t2 + vox t + xo y(t) = (1/2) ay t2 + vox t + yo
 vx(t) = ax t + vox vy(t) = ay t + voy
 We insert the known values for acceleration & initial conditions and obtain
the specific equations for the football:
 x(t) = (48)(4/5) t y(t) = - (1/2)(32) t2 + (48)(3/5) t
 vy(t) = - 32 t + (48)(3/5)

20.  We can now answer any question regarding the motion of the football. In
particular, we are interested in when it hits the ground (call this t'). We
have:
 y(t') = 0 = - 16 t'2 + (48)(3/5) t'  t' = 0, or t' = 1.8 sec.

 Hence the ball will land at x(t') = x(1.8s) = (48)(4/5)(1.8) = 69 ft
from the origin.

 We can now consider the 2nd player. His initial position (t=0) is 100 ft
from the origin, and he must reach a point 69 ft from the origin in 1.8 sec
if he is to catch the ball. Thus from the definition of average velocity,
 vave = (x2 - x1)/(t2 - t1) = (69 - 100)/(1.8)
 = - 17 ft/sec.

 The negative sign indicates that he must run toward the origin (negative
x direction).
 (Sample Problems)

21. Prepared by:
 Amonoy, Ginger A.
 Bisnar, Diwata R.
 Gonzales, Charles Jourdan V.
 Lacayanga, Moises Jerome D.
 Leoncio, Krized Noviem M.
 Santos, Ruth Margarette L.
BSCE 3-B