Scott Starks, PhD, PE
Dept. of Electrical & Computer Engineering
UTEP
Projectile Motion
1

At time t=0, an artillery shell is launched with
a vertical speed of v0 from atop a fortress that
is situated atop a mountain of height y0.
y0
1 2
y   g t  v0 t  y0
2
Projectile Motion
2
1 2
y   g t  v0 t  y0
2





y is the height (ft) of the projectile above the
ground at time t (s)
g is the acceleration constant (32 ft/s2)
v0 is the muzzle (initial) velocity (ft/s)
y0 is the height of the cannon above ground (ft)
Note: We are only considering vertical height in
this example.
Projectile Motion
3

Assume the cannon is located 672 ft above
ground level.
y0  672 ( ft )

Assume the muzzle velocity is 656 ft/s.
v0  656 ( ft / s)
Projectile Motion
4
1 2
y   g t  v0 t  y0
2
1
2
y   (32) t  (656) t  672
2
2
y  16t  656t  672 ( ft )
Projectile Motion
5
t (s)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Table
y (ft)
672
1312
1920
2496
3040
3552
4032
4480
4896
5280
5632
5952
6240
6496
6720
6912
7072
7200
7296
7360
7392
7392
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
7360
7296
7200
7072
6912
6720
6496
6240
5952
5632
5280
4896
4480
4032
3552
3040
2496
1920
1312
672
0
y (ft)
8000
7000
6000
5000
4000
3000
2000
1000
t (s)
0
0
10
20
30
40
50
Plot
Projectile Motion
6

The vertical speed (v) of the projectile
measured in (ft/s) at time (t) is described by
the equation
v  32t  v0
v  32t  656 ( ft / s)
Projectile Motion
7
t (s)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
v (ft/s)
656
624
592
560
528
496
464
432
400
368
336
304
272
240
208
176
144
112
80
48
16
-16
Table
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
-48
-80
-112
-144
-176
-208
-240
-272
-304
-336
-368
-400
-432
-464
-496
-528
-560
-592
-624
-656
-688
v (ft/s)
800
600
400
200
t (s)
0
-200 0
10
20
30
40
50
-400
-600
-800
Plot
Projectile Motion
8