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anonymous

  • 4 years ago

If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 16 ft/sec, its height after t seconds is s(t)=64+16t−16t^2. what is the maximum height the ball reaches? What is the velocity of the ball when it hits the ground (height 0)

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  1. anonymous
    • 4 years ago
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    hey notsobright can you help me with another one after please

  2. anonymous
    • 4 years ago
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    Take derivative and equate it to zero 16-32t=0 t=1/2 h=64+ 8-4 h=68

  3. anonymous
    • 4 years ago
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    Yeah sure

  4. anonymous
    • 4 years ago
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    ok

  5. anonymous
    • 4 years ago
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    what is the slope of the normal line of m2 the equation of the normal lineis y=? y=10cosx at the point (π/3,5). derivative of y is -10sinx The slope of the tangent line is m1 is -10sin(pi/3) the equation of the tangent line is y= -5sqrt(3)x+(5pi/sqrt3)+5

  6. anonymous
    • 4 years ago
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    Rephrase it it is not clear

  7. anonymous
    • 4 years ago
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    ok

  8. anonymous
    • 4 years ago
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    Find equations of the tangent line and normal line to the curve y=10cosx at the point (π/3,5). Derivative of y= -10sinx The slope of the tangent line is m1=-10sin(pi/3) The equation of the tangent line is y=-5sqrt(3)x+(5pi/sqrt3)+5 The slope of the normal line is m2=? The equation of the normal line is y=?

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