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anonymous

  • 4 years ago

using the formula h = –1/2gt^2 + v(subscrpt0)t + h(subscript0) for the height of an object under the force of gravity, with g = 32 ft/sec^2, A basketball player makes a long shot at the buzzer. He releases the ball from a height of 6 feet with an initial upward velocity of 28 ft/sec. Give an equation for the height of the ball after t seconds.

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  1. anonymous
    • 4 years ago
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    -1/2(32)t^2+28t+6=v_f

  2. anonymous
    • 4 years ago
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    not v_f its x_f

  3. anonymous
    • 4 years ago
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    y_f=y_o+V_ot-1/2gt^2 <- thats the equation you're using. This isn't a very realistic problem in my opinion because it doesn't even give you the angle he shoots the ball at. Which would really find the whole arc. Because not basketball player shoots the ball straight vertically :P

  4. anonymous
    • 4 years ago
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    its asking just for height.

  5. anonymous
    • 4 years ago
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    I understand that :) but if you wanted to find the real height you would really need to know the angle of his shot and use sin or cos. But since that's not given, the equation i gave should be correct. I don't think i'm missing anything. -1/2(32)t^2+28t+6=h

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spraguer (Moderator)
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