anonymous
  • anonymous
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−16t2. a) What is the maximum height the ball reaches? b) What is the velocity of the ball when it hits the ground (height 0)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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lalaly
  • lalaly
Set the derivative of the equation equal to zero to find the maximum height ds/dt = 16-32t= 0 t = 16/32 = .5 seconds substitute that in the equation s=64+16(0.5)-16(0.5)^2
lalaly
  • lalaly
to find part b you set s(t)=0 then solve the quadratic equation to get two values of t ignore the negative one, then u know that V(t) is derivative of s(t) so u find the derivative and substitute the time u get, to find the velocity
anonymous
  • anonymous
i dont understand how i should go about solving b

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lalaly
  • lalaly
first set s(t) =0 so u have 64+16t−16t^2=0 now solve this quadratic equations u will get two solutions for t
anonymous
  • anonymous
so i got 2.5 as the positive solution
lalaly
  • lalaly
u should get 7.5 and -0.5 are u sure u did it right?
anonymous
  • anonymous
guess not
lalaly
  • lalaly
\[t=\frac{-16 \pm \sqrt{16^2-4(-16)(64)}}{2(-16)}\]
lalaly
  • lalaly
re check ur calculations
lalaly
  • lalaly
your positive solution will be 7.5 so usualy if u say s(t) is the displacement the derivative of that is the velocity and the derivative of velocity is acceleration but here we only need the velocity so take the derivative of s(t) V=ds/dt=16-32t now substitute the t=7.5 in the above equation u will get V=16-32(7.5) And the answer is the velocity u need
anonymous
  • anonymous
ive got -224
lalaly
  • lalaly
yeah u got it right

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