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August 1, 2014

August 1, 2014

Posted by **melissa** on Tuesday, May 17, 2011 at 12:07am.

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 !!! -
**Jai**, Tuesday, May 17, 2011 at 1:50am*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~ :)

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