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Projectiles question

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    15:55

    Why is the speed at which the ball hits the ground the same as the speed at which the ball was thrown? Doesn't the variable ''g'' increase the speed of the ball so the speed at which the ball hits the ground should be greater?
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    The speed is not the same - strikes the ground at 15 and is projected at 10.6.

    Are you talking about the 10.6ms^-1 ?

    The force of gravity will only affect the vertical component of the ball's motion.

    The ball is projected horizontally so its vertical component is zero initially.
    If we find the horizontal component of its velocity using 15cos45 we can find its initial velocity.

    Also g is not a variable. We treat it as a constant.
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    (Original post by Killjoy-)
    The speed is not the same - strikes the ground at 15 and is projected at 10.6.

    Are you talking about the 10.6ms^-1 ?

    The force of gravity will only affect the vertical component of the ball's motion.

    The ball is projected horizontally so its vertical component is zero initially.
    If we find the horizontal component of its velocity using 15cos45 we can find its initial velocity.

    Also g is not a variable. We treat it as a constant.
    I mean it strikes the ground at 15cos45 = 10.6 = the launch speed right?

    Thanks
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    (Original post by Killjoy-)
    The speed is not the same - strikes the ground at 15 and is projected at 10.6.

    Are you talking about the 10.6ms^-1 ?

    The force of gravity will only affect the vertical component of the ball's motion.

    The ball is projected horizontally so its vertical component is zero initially.
    If we find the horizontal component of its velocity using 15cos45 we can find its initial velocity.

    Also g is not a variable. We treat it as a constant.
    I think I understand..

    At the instant the ball hits the ground, the vertical component of velocity is 15 m/s, and the horizontal component of velocity is 10.6 m/s right?

    The horizontal component of velocity always remains constant (as air resistance is ignored), but the vertical component of velocity increases due to the acceleration (g)?

    Is the above correct?
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    Not quite. The ball strikes the ground at an actual velocity of 15m/s at an angle of 45 degs.
    The horizontal component of this velocity is 10.6 m/s (=15 cos 45)
    As the horizontal component doesn't change and the object was initially projected horizontally, it's initial horizontal speed (and speed of projection) must have been 10.6 m/s

    The initial vertical component was zero but this has increased as a result of gravity and is now actually =15 sin 45
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    (Original post by Stonebridge)
    Not quite. The ball strikes the ground at an actual velocity of 15m/s at an angle of 45 degs.
    The horizontal component of this velocity is 10.6 m/s (=15 cos 45)
    As the horizontal component doesn't change and the object was initially projected horizontally, it's initial horizontal speed (and speed of projection) must have been 10.6 m/s

    The initial vertical component was zero but this has increased as a result of gravity and is now actually =15 sin 45
    Ah! Thank you, I understand !

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Updated: May 10, 2012
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