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Projectile motion of a plane

  1. Feb 11, 2008 #1
    1. The problem statement, all variables and given/known data

    A plane, diving with constant speed at an angle of 51.0° with the vertical, releases a projectile at an altitude of 570 m. The projectile hits the ground 4.00 s after release. (Assume a coordinate system in which the airplane is moving in the positive horizontal direction, and in the negative vertical direction. Neglect air resistance.)
    (a) What is the speed of the aircraft?

    (b) How far did the projectile travel horizontally during its flight?

    (c) What were the horizontal and vertical components of its velocity just before striking the ground?

    2. Relevant equations

    3. The attempt at a solution
    ive tried the vertical and horizontal motion equations but cannot come anywhere close to an answer
  2. jcsd
  3. Feb 11, 2008 #2


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    That's the right appraoch, use the normal laws of motion - expect that you have an initial velocity.
    The only force on the bomb is gravity vertically. So it has a constant vertical acceleration and a constant horizontal velocity.
  4. Feb 11, 2008 #3
    So if the plane's velocity is V, the initial x component is, from magical trig, is V*sin(51)

    the initial y component is V*cos(51)

    you know time, it's t=4, and you know the horizontal distance is 570m

    so -570=-1/2*g*t^2+V*cos(51)*t, remember it's negative 570 because the equation is normally d=1/2at^2+Vi*t, where d is final position - initial, and its final position is 0, initial is 570

    You know everything in that equation except V, so solve for V, and that gives you part a)

    using V you can find both Vx and Vy, and Vx lets you solve for b)

    You still know t and now know both initial velocities and accelerations involved, so you can do c
  5. Feb 11, 2008 #4


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    The 'horizontal' distance is 570m, to get the vertical impact speed just use v = u + a t
  6. Feb 11, 2008 #5
    i still cannot come up with the vertical component of velocity...
  7. Feb 11, 2008 #6
    You need to find V first, then V*cos(51)=initial y velocity

    The trig functions are reversed from your typical projectile motion because you're measuring from the vertical
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