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anonymous

  • one year ago

The height, s, of a ball thrown straight down with initial speed 32 ft/sec from a cliff 48 feet high is s(t) = –16t2 – 32t + 48, where t is the time elapsed that the ball is in the air. What is the instantaneous velocity of the ball when it hits the ground?

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  1. zepdrix
    • one year ago
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    \[\large\rm s'(t)=v(t) \qquad \leftarrow\text{this function gives us instataneous velocity}\]So we have,\[\Large\rm s(t)=-16t^2-32t+48\]Find your derivative,\[\Large\rm s'(t)=?\]

  2. zepdrix
    • one year ago
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    Oh we need a specific time t. The specific time we're looking for corresponds to the position function equaling zero. This is when the ball hits the ground, when the position, or height, is zero. It would have made more sense for them to call it h(t), height as a function of t. But they want you to get comfortable with this whole s(t), v(t), a(t) thing :) So whatev.

  3. zepdrix
    • one year ago
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    So we'll look for the time when s(t) is zero,\[\Large\rm s(t)=-16t^2-32t+48\]\[\Large\rm 0=-16t^2-32t+48\]Solve for t, Then evaluate your derivative function `at that t value`.

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