math help please !!!
posted by melissa on .
i really neeed help, i have no idea how to answer this question ! i tried everything i know!
A stone is thrown off a cliff. it reaches a maximum height of 30m after 2 seconds, then falls into the water below. it hits the water after 5 seconds
a) how high above the water is the cliff ( The maximum height is relative to water level )
b) what is the ' second zero' and what does it mean

*reposted  because i think you haven't read my previous post in this problem*
first, we solve for the initial velocity from the formula of the maximum height:
h,max = [(vo)^2]/(2g)
where
vo = initial velocity
g = acceleration due to gravity = 9.8 m/s^2
substituting,
30 = [(vo)^2]/(2*9.8)
vo = sqrt(588)
vo = 24.25 m/s
(a) note that the motion of the stone is uniformly accelerated motion, thus we can use the formula:
h = ho + (vo)*t  (1/2)*g*t^2
where
h = final height
ho = initial height
t = time
since the stone is thrown from the cliff, and making the water as reference, ho = the height of cliff, and h = 0. substituting,
h = ho + (vo)*t  (1/2)*g*t^2
0 = ho + (24.25)*(5)  (1/2)*9.8*(5^2)
0 = ho  1.25
ho = 1.25 m
(b) ..sorry but i don't know what second zero means~ ^^;
anyway, hope this helps~ :)