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S

  • 3 years ago

An object is thrown up from a height of 160 feet. Its velocity at t is v(t)=-32t+24 ft/ sec a) how high will it go? b) with what velocity will it hit the ground?

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  1. anonymous
    • 3 years ago
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    max height is at the vertex. use \(t=-\frac{b}{2a}\) to find it

  2. anonymous
    • 3 years ago
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    oh but first you need the equation. it is \[h(t)=-16t^2+v_0t+h_0\] in your case \(h_0=160\) and \(v_0=\) hmm confused. what is the initial velocity?

  3. anonymous
    • 3 years ago
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    oh it is 24, so your equation is \[h(t)=-16t^2+24t+160\] max when \(t=\frac{24}{32}=\frac{3}{4}\) replace \(t\) by \(\frac{3}{4}\) to find the max height

  4. S
    • 3 years ago
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    thank you

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