× Didn't find what you were looking for? Ask a question
Top Posters
Since Sunday
g
3
3
2
J
2
p
2
m
2
h
2
s
2
r
2
d
2
l
2
a
2
Home Q & A Board Science-Related Homework Help Physics
New Topic  
o2blea o2blea
wrote...
Posts: 124
Rep: 1 0
11 years ago
How would you solve this?
A rifle is aimed horizontally at a target 30m away. The bullet hits the target 1.9 cm below the aiming point. What are (a) the bullet's time of flight an (b) its speed as it emerges from the rifle?
Read 567 times
3 Replies

Related Topics

Replies
wrote...
#1
11 years ago
0.019=1/2(9.8)t^2,=====> t^2=0.019/4.9,
=====> t=0.06227seconds
30=V(0/.6227),=======> V=481.77m/sec
God bless you.
wrote...
#2
11 years ago
In the time the bullet traveled 30 meters, it fell 1.9 cm.  Using s = (at^2)/2 since the bullet starts at rest and falls due to the acceleration of gravity

.019 meter = 1/2(9.8) t^2       where a is 9.8 meter per second squared

solving for t gives t = 0.0623 seconds

The bullets horizontal velocity is 30 meters in .0623 seconds or  481.8 meters per second roughly 482 meters per second
wrote...
#3
11 years ago
Vx = ?
Neutral Face>? ____
..?1.9cm...........-----___
. ?............................. ...??
|?--------------30m------------?|

From linear formula
S = Vit  + ½ at ²
?.......?............?
Y = Vyt  +  ½ (-g)t ²  <== Vy = 0 (horizontal velocity only)
(- 0.019m) = 0 +½ (-9.8)t ²
(- 0.019m) = - 4.9 t ²
........________
t = ²? 0.00388       = 0.062 secs

X = Vx (t)
30m = Vx (0.062)
Vx = 30 / (0.062)    

Vx = 483.87 m/sec
New Topic      
Quick Reply
Explore
Post your homework questions and get free online help from our incredible volunteers
  837 People Browsing
 125 Signed Up Today
Start New Topic Take the Tour CoursesNew Study Tools Topics Trending Browse by Textbook
Live Chat
Related Images
  
 761
  
 234
  
 221
Your Opinion
What's your favorite math subject?
Votes: 293
Previous poll results: What's your favorite coffee beverage?