Trial Projectile in Air Projectile of Blowgun ?d x (m) ?d y (m) t (s) v x (m/s)
ID: 1487997 • Letter: T
Question
Trial
Projectile in Air
Projectile of Blowgun
?dx
(m)
?dy
(m)
t
(s)
vx
(m/s)
vf-y
(m/s)
|vf| (m/s)
?(vf)
(°)
?d
(m)
a
(m/s2)
t
(s)
Ortenzi
9.0
-0.170
0.186
48.34
-1.83
48.38
-2.16
1.38
Shawna
-0.180
0.192
46.98
-1.88
47.02
-2.29
Josh
-0.290
0.243
37.01
-2.39
37.09
-3.68
Dan M.
-0.280
0.239
37.67
-2.34
37.74
-3.55
Elijah
-0.290
0.243
37.01
-2.39
37.09
-3.68
Eric
-0.460
0.306
29.39
-3.00
29.54
-5.81
Matt S.
-0.425
0.294
30.58
-2.89
30.71
-5.37
Joe
-0.450
0.303
29.71
-2.97
29.86
-5.68
Gordon
-0.305
0.249
36.09
-2.45
36.17
-3.87
Dan J.
-0.490
0.316
28.48
-3.10
28.64
-6.18
John
-0.640
0.361
24.92
-2.54
25.17
-8.01
Ian
-0.415
0.291
30.94
-2.85
31.07
-5.25
Matt H.
-0.320
0.255
35.24
-2.51
35.33
-4.06
Jean
-0.395
0.284
31.71
-2.78
31.84
-5.00
Table 1
Calculate the acceleration of the projectile as it travels through the barrel of the blowgun.
Note: The acceleration of the projectile as it travels through the barrel of the blowgun is not 0m/s2nor -9.81m/s2.
---Please help me with what equation to use for acceleration. The information in Table 1 is all the information given, an example of solving for acceleration only needs to be shown for one trial(one student).
Trial
Projectile in Air
Projectile of Blowgun
?dx
(m)
?dy
(m)
t
(s)
vx
(m/s)
vf-y
(m/s)
|vf| (m/s)
?(vf)
(°)
?d
(m)
a
(m/s2)
t
(s)
Ortenzi
9.0
-0.170
0.186
48.34
-1.83
48.38
-2.16
1.38
Shawna
-0.180
0.192
46.98
-1.88
47.02
-2.29
Josh
-0.290
0.243
37.01
-2.39
37.09
-3.68
Dan M.
-0.280
0.239
37.67
-2.34
37.74
-3.55
Elijah
-0.290
0.243
37.01
-2.39
37.09
-3.68
Eric
-0.460
0.306
29.39
-3.00
29.54
-5.81
Matt S.
-0.425
0.294
30.58
-2.89
30.71
-5.37
Joe
-0.450
0.303
29.71
-2.97
29.86
-5.68
Gordon
-0.305
0.249
36.09
-2.45
36.17
-3.87
Dan J.
-0.490
0.316
28.48
-3.10
28.64
-6.18
John
-0.640
0.361
24.92
-2.54
25.17
-8.01
Ian
-0.415
0.291
30.94
-2.85
31.07
-5.25
Matt H.
-0.320
0.255
35.24
-2.51
35.33
-4.06
Jean
-0.395
0.284
31.71
-2.78
31.84
-5.00
Table 1
Explanation / Answer
v(t) = at v= sqrt (vx^2 + vf^2)= 68.39
s(t) = 1/2 at^2 , s(t)=d=1.38 m
at = 68.39, so a = 68.39/t
1/2 (68.39/t)t^2 = 1.38, so
t = 0.0404 s
therefore a=v/t =68.39/0.0404 =1692.82 m/s^2
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