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Homework Help: Motion in 2 directions

  1. Sep 24, 2007 #1
    1. The problem statement, all variables and given/known data
    A cannon with a muzzle speed of 1,000m is used to start an avalanche on a mountain. The target is 2,000m from the cannon horizontally and 800m above the cannon. At what angle, above the horizontal, should the cannon be fired?

    2. Relevant equations
    s^2 + b^2 = c^2 ?
    sinX = opposite/hypot.

    3. The attempt at a solution
    Is this problem as simple as simple finding the hypotenuse and taking the sin to determine the angle? I do realize gravity acts on the ball but how do i incorporate gravity into an equation to find the velocity?
  2. jcsd
  3. Sep 24, 2007 #2


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    Yes you need to take into account gravity.
    The speed horizontally is constant so you are working out how long the ball must be in the air to travel 2000m.
    If the target was the same level as the canon it is simply working out how high the ball must go to travel a certain time using s = ut + 1/2 at^2.
    Since the ball doesn't fall all the way back down - you have to work out the time to reach the top of the curve and then the time taken to free fall down to the level of the moutain.
  4. Sep 24, 2007 #3
    Unfortunately, no it's not that simple. If you shot it at the angle you calculated, that would be the same as pointing the cannon directly at your target. As you stated - gravity - you're going to miss on the low side.

    How comfortable are you with solving a system of equations? I haven't given this problem much thought, and certainly haven't tried to work it out yet, but it's possible that there are two solutions. (if you were on horizontal ground, as your angle increased from 0 to 45, (ignoring air resistance) the range would increase. Then, as the angle increase from 45 to 90, the range would decrease. For any distance out to the maximum range achieved at 45 degrees, there are two angles that will result in a hit at that distance.)
  5. Sep 24, 2007 #4
    But how do I find the times out using that equation? I can't separate the time out. Unless I split the equation into finding time for the X axis and time for Y axis, but the y axis won't work because I would have to take the sqrt root of (800 / .5*-9.8)
  6. Sep 24, 2007 #5
    I understand what you are saying dr pizza, but even so how can I acquire even one of these answers :)
  7. Sep 24, 2007 #6
    well, you could write an equation for the horizontal component and an equation for the vertical component. Both equations have time in them. Since it has to arrive simultaneously in the vertical and horizontal direction, both times need to be the same time. Thus, solve both equations for t and set them equal to one another.

    edit: if you're familiar, you can think of these as two parametric equations, one for x in terms of t, the other for y in terms of t. By eliminating the parameter t, you end up with an equation in x, y, and theta. Substitute your mountain for x and y, and solve for theta. There are multiple ways to solve this; mathematically, some methods are more difficult than others.
    Last edited: Sep 24, 2007
  8. Sep 24, 2007 #7
    I'm assuming you are referring to the same equation as mgb phys, s = ut + 1/2 at^2. I do not see how to isolate the t, or or acquire theta for that matter.
  9. Sep 24, 2007 #8
    Hard way: Yes, you can solve that equation for t.
    Quadratic formula. (Yuck! a = 1/2a, b = u, c = -s)

    That's not necessarily the easiest route though. (btw, what level of class is this? Difficult intro to physics college course? High school AP level?)

    The equation for the horizontal distance/time is a pretty simple equation to solve for t. You can always use substitution. (And, after that, there's some ugly trig involved - unless you like math - then it's beautiful trig.)
  10. Sep 24, 2007 #9
    I am in difficult intro to physics college course. I would really prefer not to use the quadratic :)
    As far as the equation for distance/ time, can I use Sx = Ucos(theta)+ 0(Because the object doesn't accelerate in the horizontal direction)?
  11. Sep 24, 2007 #10
    I think you're missing t in that equation. But, yes. s = u cos(theta) * t

    For what it's worth, I've come up with 2 solutions; both checked for me. One isn't "realistic" = while it gives a correct answer, the angle is so close to 90 degrees and the time of flight is so great that just a small error in aiming for that angle would result in a huge error.

    Also, my method required use of the trig formula: [tex]tan^2 x + 1 = sec^2 x[/tex]

    AND it required the quadratic formula with "ugly" numbers.
  12. Sep 24, 2007 #11
    but if Sx = Ux cos(theta) * t + 0,
    then Sy = O = .5*a*t^2, which doesn't make sense because a is -9.8, then solving for t it would be t^2 = Sy / .5*-9.8 (Can't take sqr root of a negative). so then what would the equation look for in the y axis?
  13. Sep 24, 2007 #12
    If someone can please explain to me how to solve this problem, I will be eternally grateful.
  14. Sep 25, 2007 #13
    If [tex]S_x=U cos(\theta) *t[/tex] Then [tex]t= \frac{S_x}{U cos(\theta)}[/tex]

    Now, write the formula for [tex]S_y[/tex] and substitute for t. Still a lot of work left.
  15. Sep 25, 2007 #14
    Sy = Usin(theta) *t - .5*g*t^2
    Not too sure how to use quadratic to solve. Bare with me.
    U sin(theta)+-Sqr Root of ((Usin)theta)^2)-(4*.5g*Sy)) / g
    If that even looks right, how am i suppose to continue with the unknown theta?
  16. Jan 21, 2010 #15
    Yf=visin(Ѳ)t-0.5gt^2 =>800=1000sin (Ѳ)t-4.9t^2
    Xf=vicos(Ѳ)t =>2=cos(Ѳ)t =>t=2/cos(Ѳ)
    => Ѳ=89.4 or 22.4 degree
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