A rifle is aimed horizontally at a target 55.0 m away. The bullet hits the target 3.00 cm below the aim point.
What was the bullet's flight time?
2006-10-03 15:56:52 · 10 answers · asked by Asperon 2 in Science & Mathematics ➔ Physics
y = 1/2 at^2
.03 = 4.9 t^2
solve for t
2006-10-03 15:59:19 · answer #1 · answered by TruthHurts 3 · 1⤊ 0⤋
1
2006-10-03 22:58:23 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Something wrong to your question. I guess, we cannot use a=9.81 as this is a downward gravity accelearation. As you shoot horizontally there must be a vector concern.
Is the question incomplete? Well, maybe you need to study the general standard of a riffle's specification to get your calculation correct.
2006-10-03 23:20:42 · answer #3 · answered by Mr. Logic 3 · 0⤊ 0⤋
s = 1/2 a t^2
3 = 1/2 9.8 t^2
t^2 = 6 / 9.8 = 0.612
t = 0.78
2006-10-03 23:09:07 · answer #4 · answered by DanE 7 · 0⤊ 0⤋
it depends on wut rifle you r using. the variable wud be the bullet speed.
using the A(sq)+B(sq)=C(sq) formula:
it wud be 55(sq)+3(sq)=C(sq), C(sq) being how far the bullet traveled
thus: 30250000cm+9cm=C(sq)
and C(sq)=30250009cm
and C= 5500.0008centimeters which is 55.000008 meters (annoying decimals)
now divide 55.000008 meters by the bullet speed shall give you the time.
2006-10-04 00:54:57 · answer #5 · answered by Frontier's 2 · 0⤊ 0⤋
What speed was the projectile fired at?
2006-10-03 23:01:45 · answer #6 · answered by Chris_Knows 5 · 0⤊ 0⤋
.03=.5*9.8*t^2
t^2=.03/4.9
t=.078 seconds
2006-10-03 22:59:55 · answer #7 · answered by bruinfan 7 · 0⤊ 0⤋
0.03=1/2(9.8)t^2
so t=0.0782 second
2006-10-03 23:21:16 · answer #8 · answered by stuart_sheng 2 · 0⤊ 0⤋
s=at, so t=s/a =0.03/9.81=0.003 sec.
2006-10-03 23:02:53 · answer #9 · answered by zee_prime 6 · 0⤊ 0⤋
.0789
2006-10-04 00:02:13 · answer #10 · answered by smart-crazy 4 · 0⤊ 0⤋