Can someone elp me with a couple of homework questions im having trouble with

In summary, if you want to aim a gun at a target 54.08 meters away, you need to aim it at an angle of 3.98 degrees radians.
  • #1

Homework Statement


A hunter aims directly at a target (on the same level) 54.08m away. If the bullet leaves the gun at a speed of 143.91 m/s then at what angle should the gun be aimed in order to hit the target? (answer in radians)


Homework Equations





The Attempt at a Solution


Vx = Vcos theta, Vx/V cos-1 = theta
im pretty sure this is not the right way to do it but i just don't know how to start or what equations to use
 
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  • #2
Involves projectile motion.
 
  • #3
so I used this equation to y=v^2y0-v2y/2g. I rearanged it using the values i had to work out vy0, which i got 32.56, i used the distance as y and v2y as 0. Then i put that into this equation vy0=v0sin theta, to work out theta. I rearranged it to vy0/v0 sin-1 = theta, i used my value for vy0 and for v0 i used the velocity and got the answer 0.23. I then changed that to radians and got the answer 3.98E-3. But that is the wrong answer...What have I done wrong??
 
  • #4
inflames829 said:
so I used this equation to y=v^2y0-v2y/2g.

Which equation is this? It does not appear to have the right dimensions. (The term on the left has units of length; the first term on the right has dimensions length^3/time^2.)


I rearanged it using the values i had to work out vy0, which i got 32.56,

I don't think you have enough information to find [itex]v_{0y}[/itex] yet. Once we find the angle [itex]\theta[/itex] that this problem is asking for, then you'll have [itex]v_{0y}=v_0 \sin\theta[/itex] (if [itex]\theta[/itex]) is the angle from the horizontal).

i used the distance as y and v2y as 0. Then i put that into this equation vy0=v0sin theta, to work out theta. I rearranged it to vy0/v0 sin-1 = theta, i used my value for vy0 and for v0 i used the velocity and got the answer 0.23. I then changed that to radians and got the answer 3.98E-3. But that is the wrong answer...What have I done wrong??

You have to keep track of motion separately in the horizontal (x) and vertical (y) direction.

In this problem, what is the displacement in the horizontal direction [itex]\Delta x[/itex]? in the vertical direction [itex]\Delta y[/itex]?

Once you have those, you may have a formula that your book derived for this special case. If so, you can calculate the angle with just that one formula.

(If you don't have this special case formula, then you can solve the problem directly using the kinematic equations for the x and y directions.)
 
  • #5
i found this equation. R= v^2 sin 2 theta/g. If i put my values in there i get 2.23 x 10 ^4, the right answer is 1.28 x 10^-2. I am pretty sure this equation is the correct one, can you please give me hints on what I've done wrong?
 
  • #6
Hi inflames829,

Yes, that is the correct question. (It is derived by setting [itex]\Delta y=0[/itex] for projectile motion with no air resistance, and so applies when that is true [itex]y_f=y_i[/itex].)

However, the answer you got is huge! The problem is asking for an angle to aim a gun, so it's looking for an answer whose magnitude is less than [itex]\pi/2[/itex] or about 1.6 or so (in radians). Did you solve your equation for theta? If so, please post what you actually put into the calculator.
 
  • #7
ive got the right answer now. I realize what i did wrong. Thanks for the help though
 

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