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Math Help - Applications for Derivatives (velocity)

  1. #1
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    Applications for Derivatives (velocity)

    *insert long descriptive paragraph about a volcano*
    Paraphrase: With regards to a shot of lava at a point 1900 ft shooting straight into the air, what was the lava's exit velocity in feet per second? in miles per hour?

    [HInt: If v0 is the exit velocity of a particle of lava, its height t seconds later will be s= v0t - 16t^2 feet. Begin by finding the time at which ds/dt= 0. Neglect air resistance]

    I assume v0 is just another term for the first velocity.

    ds/dt= v0 - 32t

    v0 - 32t = 0

    v0 = 32t

    That seems unhelpful.

    What now? or did I misunderstand?


    Thanks!
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  2. #2
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    Hello, Truthbetold!

    With regards to a shot of lava at a point 1900 ft shooting straight into the air,
    what was the lava's exit velocity in feet per second? in miles per hour?

    Hint: If v_o is the exit velocity of a particle of lava,
    its height t seconds later will be: .  s \:= \:v_ot - 16t^2 feet.
    Begin by finding the time at which \frac{ds}{dt} \,= \,0

    I assume v_o is just another term for the first velocity. .
    . . . Yes!

    \frac{ds}{dt} \:= \:v_o - 32t \:=\:0\quad\Rightarrow\quad v_o \:=\:32t
    . . .
    You were well on your way . . .

    You found that the lava reached maximum height at: . t \,=\,\frac{v_o}{32} seconds.

    Then the maximum height is: . s \;=\;v_o\left(\frac{v_o}{32}\right) - 16\left(\frac{v_o}{32}\right)^2 \;=\;\frac{v_o^2}{64} feet


    But we are told that the maximum height is 1900 feet.

    . . So we have: . \frac{v_o^2}{64} \:=\:1900

    Got it?

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