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Bob
Bob asked in Science & MathematicsPhysics · 2 years ago

I have one more physics question?

At what velocity must you launch a pumpkin from a cannon such that when your friend standing at the top of a 565m deep canyon can catch it a 0m/s velocity?

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  • oubaas
    Lv 7
    2 years ago
    Favorite Answer

    Vy = √2gh = √1130*9.806 = 105.26 m/sec

  • oldschool
    Lv 7
    2 years ago

    This is a very poorly written question. The only way the pumpkin would have both vertical and horizontal velocity = 0 is at the peak height of a perfectly vertical trajectory.

    Vf² = Vi²+2*a*h = 0² = Vi² - 2*9.8*565 

    Vi² = 19.6*565 = 105²

    Vi = 105m/s

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