9 pm, and still dont understand

  • Thread starter megg0230
  • Start date
In summary, the problem involves a rifle bullet being fired at an angle of 30 degrees below the horizontal with an initial velocity of 800 m/s from a cliff 80 m high. Using kinematic equations and resolving the initial velocity into X and Y coordinates, we can determine the displacement of the bullet from the base of the cliff. The 1st equation can be used to find the time and the 2nd equation can be used to find the horizontal displacement. Taking into account the angle and height, we can solve for the final answer.
  • #1
megg0230
2
0

Homework Statement


A rifle bullet is fired at an angle of 30 degrees below the horizontal with an initial velocity of 800 m/s from the top of a cliff 80 m high. How far from the base of the cliff does it strike the level of ground below
g=9.8
initial velocity- 800 m/s
y=80 m


Homework Equations


People i ask say..
R=(initial velocity squared x sin(2 theta)
g




The Attempt at a Solution


i did
800^2 x sin2(30)
9.8

i get 56556.76106-which seems way to high
Then i did Soh Cah Toa
Sin=Opp/Adj
56556.76106/80
and got 706.9595133

I really just don't understand what i am doing, and need help. bad. this is due tomorrow.
 
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  • #2
Hello welcome to Physicsforums! Hope you will learn here and also teach us what you know!

For the problem:
(1)You must resolve the initial velocity into their X and Y coordinates using trigonometry.
(2)Then use kinematic equations to solve the problem.
 
  • #3
I don't even know any of that.
We never learned Kinematic equations.
Im very slow at this, and am now regretting taking this class.
 
  • #4
Hey don't despair! I can teach you what I know. I am learning this myself.

Kinematic Equations:
S=Vot+0.5at^2
S=Vft-0.5at^2
S=((Vo+Vf)/2)*t
Vf=Vo + at
Vf^2=Vo^2 + 2aS

S=Displacement in metres
Vo=Initial Velocity in metres/second
Vf=Final Velocity in metres/second
a=acceleration in metres/second squared
t=time in seconds

For all kinematic problems, write your 5 variables down and write the known values in.

Velocity is a vector quantity. It has magnitude and DIRECTION. direction is usually indicated by an angle as in your problem.

Velocity of the bullet is 800m/s angle 30degrees below the horizontal. Draw a picture and see if you can resolve the velocity into X and Y coordinates.

Have a look at this site:
http://www.glenbrook.k12.il.us/gbssci/phys/Class/vectors/u3l2c.html

Then we will see if we can solve the problem. Please come back.
 
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  • #5
megg0230 said:
I don't even know any of that.
We never learned Kinematic equations.
Im very slow at this, and am now regretting taking this class.
http://www.mathlinks.ro/Forum/latexrender/pictures/b/0/2/b02ab4a44802a89197dc4fcc4947b1ec33ebdf9d.gif
 
Last edited by a moderator:
  • #6
Here is a better site.

http://id.mind.net/~zona/mstm/physics/mechanics/curvedMotion/projectileMotion/generalSolution/generalSolution.html
 
Last edited by a moderator:
  • #7
use the 2nd equation to find time

keep in mind the bullet is being fired DOWN, so your y value must be negative

when you find time, disregard the negative value

then use the 1st equation to find x
 

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