Question

If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 64 ft/sec, then what is the maximum height the ball reaches?
What is the velocity of the ball when it hits the ground?

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
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
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
u should get 7.5 and -0.5

Answer 2

I plugged in 7.5 for maximum height and it doesnt work

Answer 3

@im.celibate

Answer 4

Oh. That is weird because I got the answer right just like 2 hours ago..

Answer 5

i plugged it online it does not wrk

Answer 6

oh, that sucks

Answer 7

wow thanks

Answer 8

Use the formula:
v^2 = u^2 + 2as
Where the final velocity is 0 (maximum height: think of it...) initial velocity = 64, acceleration = 32.2 and s = displacement (your maximum height)

Answer 9

So:
0^2 = 64^2 + 2 x 32.2 x s
Now solve for s