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Thread: Projectile motion

  1. #1
    Mtl
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    Projectile motion

    I know Projectile motion is physics in nature but i am having problems with the mathematical mechanics and any help would be greatly appreciated,

    A fighter pilot blows up a artillery dump. Diving at an angle of 53degrees below the horizon, the pilot releases a bomb at an altitude of 750m. The bomb hits the target 5.0s after being released.
    What is the velocity of the bomber?

    So i enter this into the equation d= vt - (1/2)gt^2 yet i get a negative number which to me doesnt make sense.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by Mtl View Post
    I know Projectile motion is physics in nature but i am having problems with the mathematical mechanics and any help would be greatly appreciated,

    A fighter pilot blows up a artillery dump. Diving at an angle of 53degrees below the horizon, the pilot releases a bomb at an altitude of 750m. The bomb hits the target 5.0s after being released.
    What is the velocity of the bomber?

    So i enter this into the equation d= vt - (1/2)gt^2 yet i get a negative number which to me doesnt make sense.
    Did you consider that the equation for $\displaystyle V_x$ and $\displaystyle V_y$ are differnt...I hope I am not making a fool out of myself lol
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  3. #3
    Mtl
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    "Did you consider that the equation for and are differnt"

    Yes. I should have defined all my variables a little better. The d = distance in the y direction (in metres), the v = initial velocity of the bomber (metres/second), t = time (seconds)

    "...I hope I am not making a fool out of myself lol"

    It was a valid question, thanks for your help.
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  4. #4
    Senior Member topher0805's Avatar
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    Quote Originally Posted by Mtl View Post
    "Did you consider that the equation for and are differnt"

    Yes. I should have defined all my variables a little better. The d = distance in the y direction (in metres), the v = initial velocity of the bomber (metres/second), t = time (seconds)

    "...I hope I am not making a fool out of myself lol"

    It was a valid question, thanks for your help.
    I think that what he is trying to say is that you have set up you equation wrong. You can not just do it in one step. This is vector dynamics so velocity and acceleration have direction.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mtl View Post
    I know Projectile motion is physics in nature but i am having problems with the mathematical mechanics and any help would be greatly appreciated,

    A fighter pilot blows up a artillery dump. Diving at an angle of 53degrees below the horizon, the pilot releases a bomb at an altitude of 750m. The bomb hits the target 5.0s after being released.
    What is the velocity of the bomber?

    So i enter this into the equation d= vt - (1/2)gt^2 yet i get a negative number which to me doesnt make sense.
    Define a coordinate system where the origin is on the ground directly below the plane when the bomb is dropped. I'm taking +x in the direction the plane is traveling and +y straight up.

    We know for the bomb that
    $\displaystyle x_0 = 0~m$....$\displaystyle x = $?
    $\displaystyle v_{0x} = v_0~cos(53)$....$\displaystyle v_x = $?
    $\displaystyle a_x = 0~m/s^2$

    $\displaystyle y_0 = 750~m$....$\displaystyle y = 0~m$
    $\displaystyle v_{0x} = -v_0~sin(53)$....$\displaystyle v_y = $?
    $\displaystyle a_x = -9.8~m/s^2$

    where $\displaystyle x, v_x, y, v_y$ are the positions, etc. at time t = 5.0 s, and $\displaystyle v_0$ is the speed the plane is moving at as it drops the bomb.

    We have three main equations:
    $\displaystyle x = x_0 + v_{0x}t$
    $\displaystyle y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$
    $\displaystyle v_y = v_{0y} + a_yt$

    See if that helps. (The basic idea is that we have the motion equations independent of each other in each coordinate direction.)

    -Dan
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