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A stone thrown horizontally from a height of 7.8 m hits the ground at a distance of 12.0 m. Calculate the speed of the ball as it hits the ground.

I was able to find the initial speed of the ball, which is 9.52 m/s. However, I'm unsure about whether I need to use this number to get the speed as it hits the ground.

  • Physics -

    The speed which which it was thrown is 12.0 m/(fall time) = The fall time is
    T = sqrt (2H/g) = 1.262 s
    Vox = 12.0/1.262 = 9.51 m/s. Close enough. You WILL need that number to finish the problem. It remains the horizontal component of velocity until it hits the ground.
    The vertical component of velocity when it hits the ground is
    Vyfinal = sqrt (2gH) = 12.36 m/s

    When it hits the ground, the speed is
    Vfinal = sqrt[Vox^2 + Vyfinal^2] =?

  • Physics -

    15.6 m/s. I understand now, thanks a bunch.

  • Physics -

    Yes, that's great explain. Also, you be able to use suvat formulae in steps to work out time and then vox and then resultant velocity.

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