A projectile is an object upon which the only force acting is gravity. There are many examples of projectiles. An object dropped from rest is a projectile as long as that the influence of air resistance is negligible. An object that is thrown vertically upward is also a projectile provided that the influence of air resistance is negligible. And an object which is thrown upward at an angle to the horizontal is also a projectile as long as that the influence of air resistance is negligible. A projectile is any object that once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.

1. ProjectileMotion
By Prof. Liwayway Memije-Cruz

2. Projectile
isabody
launched at an
anglefollowing
acurvaturecalled
thetrajectory
path.
an object upon
which theonly
forceacting is
gravity.

3. Examplesof Projectiles
Thereareavariety of examplesof projectiles. An object
dropped from rest isaprojectile(provided that theinfluence
of air resistanceisnegligible). An object that isthrown
vertically upward isalso aprojectile(provided that the
influenceof air resistanceisnegligible). And an object which
isthrown upward at an angleto thehorizontal isalso a
projectile(provided that theinfluenceof air resistanceis
negligible). A projectileisany object that once pro jected or
dropped continuesin motion by itsown inertia and is
influenced only by thedownward forceof gravity.

5. Regardlessof whether
aprojectileismoving
downwards, upwards,
upwardsand
rightwards, or
downwardsand
leftwards, thefree-
body diagram of the
projectileisstill as
depicted in thediagram
at theright.

6. First caseof trajectoryFirst caseof trajectory
Thetrajectory isahalf parabola

7. Newton’s First Law of Motion: Newton’s Law ofNewton’s First Law of Motion: Newton’s Law of
IneInertiartia

9. Second case of trajectorySecond case of trajectory
Trajectory isafull parabola

10. A projectileisan object upon which theonly forceis
gravity. Gravity actsto influencethevertical motion
of theprojectile, thuscausing avertical acceleration.
Thehorizontal motion of theprojectileistheresult of
thetendency of any object in motion to remain in
motion at constant velocity. Dueto theabsenceof
horizontal forces, aprojectileremainsin motion with
aconstant horizontal velocity. Horizontal forces
are not required to keep aprojectilemoving
horizontally. Theonly forceacting upon aprojectile
isgravity!

11. Problem 1: An object islaunched at avelocity of 20 m/sin a
direction making an angleof 25° upward with the
horizontal.
a) What isthemaximum height reached by theobject?
b) What isthetotal flight time(between launch and touching
theground) of theobject?
c) What isthehorizontal range(maximum x aboveground)
of theobject?
d) What isthemagnitudeof thevelocity of theobject just
beforeit hitstheground?

13.  a) What isthemaximum height reached by theobject?
Theformulasfor thecomponentsVx and Vy of thevelocity and componentsx and y
of thedisplacement aregiven by
Vx = V0 cos(θ) Vy = V0 sin(θ) - g t
x = V0 cos(θ) t y = V0 sin(θ) t - (1/2) g t2
In theproblem V0 = 20 m/s, θ = 25° and g = 9.8 m/s2
.
Theheight of theprojectileisgiven by thecomponent y, and it reachesits
maximum valuewhen thecomponent Vy isequal to zero. That iswhen theprojectile
changesfrom moving upward to moving downward.(seefigureabove) and also the
animation of theprojectile.
Vy = V0 sin(θ) - g t = 0
solvefor t
t = V0 sin(θ) / g = 20 sin(25°) / 9.8 = 0.86 seconds
Find themaximum height by substituting t by 0.86 secondsin theformulafor y
maximum height y (0.86) = 20 sin(25°)(0.86) - (1/2) (9.8) (0.86) 2
= 3.64 meters

14. b) Thetimeof flight istheinterval of timebetween when
projectileislaunched: t1 and when theprojectiletouchesthe
ground: t2. At t = t1 and t = t2, y = 0 (ground). Hence
V0 sin(θ) t - (1/2) g t2
= 0
Solvefor t
t (V0 sin(θ) - (1/2) g t) = 0
two solutions
t = t1 = 0 and t = t2 = 2 V0 sin(θ) / g
Timeof flight = t2 - t1 = 2 (20) sin(θ) / g = 1.72 seconds.

15. c) In part c) abovewefound thetimeof
flight t2 = 2 V0 sin(θ) / g. Thehorizontal
rangeisthehorizontal distancegiven by
x at t = t2.
range= x(t2) = V0 cos(θ) t2 = 2 V0 cos(θ)
V0 sin(θ) / g = V0
2
sin(2θ) / g = 202
sin
(2(25°)) / 9.8 = 31.26 meters

16. d) Theobject hitstheground at t = t2 = 2 V0 sin(θ) / g (found in part b
above)
Thecomponentsof thevelocity at t aregiven by
Vx = V0 cos(θ) Vy = V0 sin(θ) - g t
Thecomponentsof thevelocity at t = 2 V0 sin(θ) / g aregiven by
Vx = V0 cos(θ) = 20 cos(25°) Vy = V0 sin(25°) - g (2 V0 sin(25°) / g) = -
V0 sin(25°)
ThemagitudeV of thevelocity isgiven by
V = √[ Vx
2
+ Vy
2
] = √[ (20 cos(25°))2
+ (- V0sin(25°))2
] = V0 = 20 m/s

17. Problem 2: A projectileislaunched from
point O at an angleof 22° with an initial
velocity of 15 m/sup an inclineplanethat
makesan angleof 10° with thehorizontal.
Theprojectilehitstheinclineplaneat point
M.
a) Find thetimeit takesfor theprojectileto hit
theinclineplane.
b) Find thedistanceOM.