Motion in space: velocity and acceleration

  1. 1. The problem statement, all variables and given/known data

    A gun has a muzzle speed of 80 meters per second.
    What angle of elevation should be used to hit an object 150 meters
    away? Neglect air resistance and use g = 9.8m/sec2 as the
    acceleration of gravity.

    2. Relevant equations

    x = (vcos//theta)t
    y = (vsin//theta)t - 1/2gt^2
    g = 9.8 m/sec^2

    3. The attempt at a solution

    i substituted the given values:
    x should be 150 and y should be 0 right?

    150 = (80 cos\\theta)t
    0 = ( 80sin\\theta)t - 4.9t^2

    i can't figure out how to eliminate a variable or how to begin to solve for //theta
    i solved for t using the "x" equation ( t = 15/8cos\\theta) and put that into the "y" equation
    and i got this...
    0 = 150tan//theta - 73.5/64(sec\\theta)^2
    where do i go from here to find \\theta?
  2. jcsd
  3. One thing I can recommend is to use symbols until you have a quadratic to solve and then plug in numbers. It's easier to spot mistakes and somebody trying to help you can follow the equations without having to do arithmetic in their heads. It always helps me but I hate arithmetic ;)

    I wonder if you could substitute something for [tex]sec^{2} \theta[/tex]. There might be a trig identity. It might even involve [tex]tan^{2} \theta[/tex] which would be sweet 'cause then you'd have a quadratic for the tangent.
  4. oh (sec\\theta)^2 = 1 + tan//theta^2 !!
    i'll make //theta = Q since it's kind of a hassle to type it all the time...
    so far its
    0 = 150tanQ - 73.5/64(1 + (tanQ)^2)
    and i did the quadratic formula but that gives me values when tanQ = 2 different long numbers that i got
    then i untangent it ( or tan^-1) and both numbers aren't close to the right answer
    ...what am i doing wrong?
    p.s. i hate arithmetic too
  5. You're doing too much arithmetic. Post the quadratic you get up here without any numbers in it (Well, 2 in the exponent is acceptable) and somebody will check it out for you. g=g, v=v, x=x, etc.

    For example, t = (x/v)*secQ
  6. with no numbers???

    i don't understand but heres my equation at first though...:
    x = tanQ
    0 = 150x - 73.5/64(1 + x^2)

    0 = 150x - 73.5/64 - (73.5/64)x^2
    to ( multiplied everything by -1)
    0 = (73.5/64)x^2 - 150x + 73.5/64

    then i put it in the quadratic formula ( x = [-b +/- sqrt(b^2 - 4ac)] / 2a)
    with x = tanQ

    tanq = [150 +/- sqrt(22500 - 4(73.5/64)^2)] / 2(73.5/64)

    simplified a bit....

    tanq = [150 + sqrt(22500 - ~5. 3)] / 73.5/32)


    tanq = [150 - sqrt(22500 - ~5. 3)] / 73.5/32)

    after i get the value of both right hand sides,
    i get the tan^-1 of those values (in radians cuz theanswr is in radians)
    and they both DON'T get me this answer: 0.115878293158862 radians
    am i doing something wrong?
    Last edited: Oct 3, 2009
  7. Well if x = (vcosQ)t and y = (vsinQ)t - (g/2)t2, then solve x for t and substitute that into y without replacing any of the variables with the numbers that you want too. Simplify it to get a quadratic in tanQ. If you want, at this point, put your formula up here and I will see if it's right. After you have done this, you can safely plug in the values that you want for y, x and v and solve the equation for tanQ.

    This way, all the arithmetic errors can be contained to one part of the calculation and it cuts down the arithmetic involved over all, hence making a mistake less likely.
  8. oh ok
    t = x /(vcosQ)
    y = (vsinQ)[x /(vcosQ)] - (g/2)[x /(vcosQ)]^2

    y = xsinQ/ cosQ - g/2 (x/v) (1/cosQ)^2

    y = x(tanQ) - g/2 (x/v) (1 + (tanQ)^2)
    y = g/2(x/v)tanQ^2 + tanQx + g/2(x/v)
    that's right, right?

    then i put it into: tanQ = [-x +/- sqrt( x^2 - 4(g/2[x/v]^2))] / 2(g/2(x/v))
    after that: tan^-1(tanQ)
    that's right too right?
  9. x and v get squared in the t2 term and you have a sign error in distributing
    -g/2 (x2/v2).

    The best thing is to substitute the numbers in just before you solve the quadratic if you are going to use the quadratic formula. It's much less awkward that way.
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?