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~ :)

I understand that you're having trouble answering the given question. Don't worry, I'm here to help you out!

Let's break down the problem step by step:

a) To find the height of the cliff, we need to consider the motion of the stone. We know that the stone reaches a maximum height of 30m after 2 seconds and hits the water after 5 seconds. This information implies that the stone goes up to a certain height and then falls back down.

During the upward motion, the stone experiences a constant deceleration due to gravity until it reaches its maximum height. Then, during the downward motion, it experiences the same constant acceleration due to gravity.

To solve this, we can use the following kinematic equation:
h = v₀t + (1/2)gt²,

where:
h = height (relative to the water level)
v₀ = initial velocity (which is zero when the stone is thrown upward)
t = time taken to reach the highest point (2 seconds in this case)
g = acceleration due to gravity (approximately 9.8 m/s²)

Using this equation, we can calculate the height:
h = 0 + (1/2) × 9.8 × (2)²
h = 0 + (1/2) × 9.8 × 4
h = 0 + 19.6
h = 19.6 meters

Therefore, the height of the cliff above the water level is 19.6 meters.

b) The "second zero" refers to a concept in the stone's motion. To understand it, we need to consider the time taken for the stone to hit the water.

In this case, the stone reaches its maximum height of 30 meters after 2 seconds and hits the water after a total time of 5 seconds. The "second zero" occurs when the stone hits the water. It represents the moment when the stone's displacement (height above the water level) becomes zero.

So, in this context, the "second zero" simply refers to the time at which the stone hits the water, which is 5 seconds in this case.

I hope this explanation helps you understand how to approach this problem. Let me know if there's anything else I can assist you with!