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)?

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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
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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
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
i dont understand how i should go about solving b

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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
so i got 2.5 as the positive solution
u should get 7.5 and -0.5 are u sure u did it right?
guess not
\[t=\frac{-16 \pm \sqrt{16^2-4(-16)(64)}}{2(-16)}\]
re check ur calculations
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
ive got -224
yeah u got it right

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