Question

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)?

Answer 1

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

Answer 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

Answer 3

i dont understand how i should go about solving b

Answer 4

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

Answer 5

so i got 2.5 as the positive solution

Answer 6

u should get 7.5 and -0.5 are u sure u did it right?

Answer 7

guess not

Answer 8

\[t=\frac{-16 \pm \sqrt{16^2-4(-16)(64)}}{2(-16)}\]

Answer 9

re check ur calculations

Answer 10

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

Answer 11

ive got -224

Answer 12

yeah u got it right