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SAMPLE PROBLEMS: 111-SET #4 PROJECTILE MOTION (2-D MOTION PROBLEMS):
1) 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?
Solution: 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:
vo
x
6.4 m
a
x(t=0) = xo = 0 ; y(t=0) = yo = 0
y
vx(t=0) = vox = 350 m/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
vx(t) = ax t + vox
y(t) = (1/2) ay t2 + voy t + yo
vy(t) = ay t + voy
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.
2) 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?
Solution: 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.
y
a
vo
37
x
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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)
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).
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.
Solution: 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
yo = 0
vox = vo cos 60 = (.5) vo
voy =
y
a
vo
60
x
ax = 0 .
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
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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:
y
48 ft
vo
x(t') = 80 = (.5) vo t'
x
y(t') = 48 = -(1/2)(32)t'2 + (.866)vo t' .
80 ft
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=
( v x ) 2  ( v y ) 2  (33.6) 2  (17.9) 2 = 38.1 ft/sec.
tan  = (17.9)/(33.6) 
33.6
vx
17.9
vy

v
= 28o (below hor. as shown)
4) A basketball player releases the ball 7 ft above the floor when he is 30 ft from the basket. The ball goes
through the rim of the basket (which is 10 ft above the floor) 1.5 seconds after release. Find the initial
velocity, and the maximum height above the floor reached by the ball.
Solution: The setup for a projectile motion problem is: (1) figure drawn;
(2) CS chosen; (3) initial conditions determined; (4) acceleration
identified. (Note we have selected a CS with origin (y=0) on floor.) The
initial conditions are:
xo = 0 ; vox = vo cos  ; yo = 7 ft ;
y vo

a
7 ft
voy = vo sin  .
x
The acceleration in the problem is:
ax = 0
ay = - 32 ft/s2 .
Inserting these values into the general equations of motion in 2-dimensions, we have:
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x(t) = vo cos  t ;
y(t) = - (1/2)(32) t2 + (vo sin  ) t + 7 ;
At t = 1.5 seconds we have:
x(1.5s) = 30 = vox (1.5)
vox = vo cos  = 20 ft/sec;
Thus:
vy(t) = - 32 t + vo sin  .
y(1.5s) = 10 = - 16(1.5)2 + voy(1.5) + 7
voy = vo sin  = 26 ft/sec.
The magnitude & direction of the initial velocity is then:
vo =
(20) 2  (26) 2 = 32.8 ft/sec; tan  = (26)/(20)  = 52.4o
(That is, 52.4o above the horizontal).
The maximum height above the floor occurs at a time t' when vy(t') = 0. Hence:
vy(t') = 0 = - 32 t' + 26  t' = .866 sec.
Then
y(0.866sec) = - 16 (.866)2 + (26)(.866) + 7 = 17.52 ft.
5) You wish to fire an anti-aircraft shell to intercept an enemy plane flying towards you at 600 mph and an
altitude of 42,000 ft. If the plane is initially 20 miles away, and your artillery piece has a muzzle velocity of
2000 ft/sec, how long to you have to adjust the piece to an angle of 60o and fire?
Solution: We have a projectile motion problem (as far as the shell is
concerned). Hence we have drawn a figure, chosen a CS, have written
down the initial conditions (initial position & velocity) of the shell (at
t=0).
y
a
vo
60
xo = 0 vox = vo cos 60 = (.5)(2000) = 1000 ft/sec
ax = 0 .
x
yo = 0 voy = vo sin 60 = (.866)(2000) = 1800 ft/sec ay = - 32 ft/s2 .
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
shell:
x(t) = 1000 t ;
y(t) = - (1/2)(32) t2 + 1800 t ;
vy(t) = - 32 t + 1800 .
We can now answer any question regarding the motion of the shell. One question that might occur to us is:
Can we hit a plane at 42,000 ft.? Let's determine the maximum height to which the shell will rise.
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vy(t') = 0 = -32 t' + 1800  t' = 56.3 sec.
Then: ymax = y(t') = -(1/2)(32)(56.3)2 + 1800 (56.3) = 50,625 ft.
We can reach the elevation of the plane. The next question is "When will the shell be at the plane's
elevation (42,000 ft)?
y(t') = 42,000 = - 16 t'2 + 1800 t' --> t'2 - 112.5 t'+ 2625 = 0
We have a quadratic equation to solve. Using the quadratic formula we have:
t' =
 b  b2  4a c
2a
=
(112.5)  (112.5) 2  4(2625)
2
This yields the results: t'1 = 33.03 sec; t'2 = 79.47 sec.
At the 1st time the shell will be a distance x(t'=33.03) = 33,030 ft (6.256 mi) from where it was fired. At
the 2nd time x(t'= 79.47) = 79,470 ft (15.05 mi).
But the plane travelling at 880 ft/sec will cover a distance of 13.245 mi in 79.47 seconds. Hence, we can't
hit the plane with the shell on the way down.
We illustrate in the figure the various distances
involved. xh = 6.256 mi is position where we
expect to hit the plane; xo = 20 mi is initial
position; xf = ?? is position of plane when we fire
the shell. Since the plane will travel a distance of
(880)(33.03)/(5280) = 5.505 mi while the shell is
in the air, then xf = 6.256 + 5.505 = 11.76 mi
away.
y
vo
xh
xf
xo
Now the question is, "How long do we wait till the plane reaches xf? Taking the difference (20 - 11.76) and
dividing by the planes speed, gives us 49.44 seconds. There is one other 'minor' question we should ask
ourselves. Where will the plane drop its bomb??? If it will release its bomb at a distance greater than the
11.76 miles then we better get the heck out of here!!! A falling bomb will have equations of motion given
by: (using CS with y=0 the ground)
xB(t) = 880 t
yB(t) = -(1/2)(32) t2 + 42,000 .
Hence, the time to 'fall" is:
yB(t') = 0 = -16 t'2 + 42,000  t' = 51.23 sec.
In this time the bomb travels a horizontal distance of (880)(51.23)/5280 = 8.54 mi. Thus the plane plans to
release its bomb when it is 8.54 miles from us. But we know, the plane will never reach this point. It will be
shot down 11.76 miles away!