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math help please !!!

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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

  • math help please !!! - ,

    *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)
    vo = initial velocity
    g = acceleration due to gravity = 9.8 m/s^2
    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
    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~ :)

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