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Magnitude of the Initial Velocity

  1. Sep 13, 2003 #1
    Given: g = 9.8 m/s^2

    A cannon sends a projectile towards a target a distance 1250m away. The initial velocity makes an angle 40 degrees with the horizontal. The target is hit.

    What is the magnitude of the initial velocity?
    How high is the highest point of the trajectory?
    How long does it take for the projectile to reach the target?

    I tried to find the time by using the following formula:

    t = sqrt (2d/g)

    and then I used

    x = x(initial) +v(initial)t + 1/2gt^2

    and I get a decimal answer...
    And then I took the distance and divided it by 2 ad I got 625 and did tan40 = h/625 but when i entered in the answer, it's wrong.
    For the 3rd Q, I used the time that I found earlier from the formula, but that's incorrect too.

    Any help would be appreciated.
  2. jcsd
  3. Sep 13, 2003 #2
    note that you can use the following formula, I will use 'u' for the symbol for initial velocity

    v = at + u

    this can model the particle's verticle component

    u = q sin(40) ... where q is the initial speed

    if we want to find the time it takes to go up to its max point i.e. its turning point

    v = 0

    so, 0 = -9.8t + q sin(40)
    so, t = q sin(40) / 9.8

    that is the time to go up which is equal to the time to come back down again, so the total time is (I'll call the total time T)

    T = 2t
    T = q sin(40) / 4.9

    now we revert to the horizontal component of the situation using the formula

    s = s0 + u*t + 0.5*a*t^2

    s0 = 0
    u = q cos(40)
    t = T = q sin(40) / 4.9
    a = -9.8
    s = 1250

    1250 = q cos(40) q sin(40) / 4.9 - 0.5*9.8*(q sin(40) / 4.9)^2
    1250 = q^2 cos(40) sin(40) / 4.9 - q^2 sin^2(40) / 4.9
    q^2 = 1250 / [ cos(40) sin(40) / 4.9 - sin^2(40) / 4.9 ]
    q = √( 1250 / [ cos(40) sin(40) / 4.9 - sin^2(40) / 4.9 ] )
    q ≈ 278.0442 m/s

    so the initial speed of the object is approxomately 278.0442 m/s

    remember that

    T = q sin(40) / 4.9
    T = 278.0442 sin(40) / 4.9
    T ≈ 36.4742 s

    so it takes approxomately 36.4742 seconds to reach the target. Now for the last part of the question we use the formula

    s = s0 + u*t + 0.5*a*t^2

    we will apply this to the verticle component of the situation, the time to go up to the maximum is just half of the total time of flight so

    s0 = 0
    t = 36.4742 / 2
    u = 278.0442 sin(40)
    a = -9.8

    s = 278.0442 sin(40) 36.4742 / 2 - 4.9 (36.4742 / 2)^2
    s ≈ 1629.696 m

    so the maximum height the projectile will reach is approxomately 1629.696 m.

    Thats the problem done! :smile:
  4. Sep 14, 2003 #3


    User Avatar
    Science Advisor

    Your mistake was initially using "t = sqrt (2d/g)". That's the same as d= (g/2)t^2 which is the distance an object will FALL under gravity, starting from rest.

    In this case, if the initial speed is v, then the initial horizontal component of speed is v cos(40) and the initial vertical component is v sin(40).

    There is no acceleration horizontally and the acceleration due to gravity, -g, vertically. If we use x for horizontal distance and y for vertical distance,

    x= v cos(40) t and y= v sin(40)t- g/2 t^2.

    The 1250 m distance to the target is horizontal, of course, so you can solve x= v cos(40) t= 1250 to find the time, t, when the projectile hits the target. That will, of course, depend on v.

    Now, use the fact that y must be 0 again to hit the target at that time: y= v sin(40)t- g/2 t^2= 0 with the value of t you just found to get an equation to solve for v.
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