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anonymous

  • one year ago

Awheeled car with a spring-loaded cannon fires a metal ball vertically. If the vertical initial speed of the ball is 2.0 m/s as the cannon moves horizontally at a speed of 0.55 m/s , how far from the launch point does the ball fall back into the cannon?

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  1. anonymous
    • one year ago
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    Hey Rachel, This is a fairly easy one so i'm going to try and make you fully understand it. lets first under stand the problem. We know the 'projectile's' initial velocity which is 2m/s, we can call this the 'projectiles (y) component' because it is in the (y) or upwards direction. we also know the forward velocity ((x) component) that is 0.55 m/s. now lets visualize this while also listing our unknowns; |dw:1443406998862:dw| The hard part of the question is solving for the amount of time taken. The time taken is completely dependent on its initial y velocity (2m/s). (the reason why the velocity in the 'x' direction doesn't matter is the same reason why if you through w ball up in a moving buss it comes right back to your hand as if you were not moving). Now visualize a ball being thrown straight up, it will keep on slowing down (decelerating) because gravity is acting on it (10m/s^2). lets draw a graph. |dw:1443408132615:dw| here, the gradient is the velocity / time (change in velocity) so gradient=rise/run =4/t if we change the subject of the formula we get t=rise/gradient=4/10= 0.4seconds. all we need to do now is multiply the time it took for the ball to came back into the forward velocity. distance=speed x time = 0.55 x 0.4 = 0.22m ( the answer)

  2. anonymous
    • one year ago
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    Did this help you? Because if not please ask me anything you did not understand

  3. IrishBoy123
    • one year ago
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    what a great few posts!! than you @zakisista IMHO, you should always draw these before solving. and the v-t graph was the icing on the cake.

  4. anonymous
    • one year ago
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    Thank you!

  5. anonymous
    • one year ago
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    @IrishBoy123

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