Solve for Height of Cliff After Diver Jumps Off at 2.2m/s

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SUMMARY

The problem involves calculating the height of a cliff from which a diver jumps horizontally at a speed of 2.2 m/s and reaches the water in 2.5 seconds. The correct formula to use is Y = Yo + Vyo*t - 0.5gt^2, where Y is the final height (0 at water level), Yo is the initial height, Vyo is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (9.8 m/s²), and t is the time (2.5 s). The correct calculation yields a height of 30.625 meters, not 25 meters as initially calculated by the user.

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Homework Statement


A diver running 2.2 m/s dives out horizontally from the edge of a vertical cliff and 2.5 s later reaches the water below. How high was the cliff?


Homework Equations


Y=Yo+Vyo*t-.5gt^2

I have Y as being 0 (the point when they hit the water) and velocity of y at time zero being 2.2 m/s.


The Attempt at a Solution


I used the equation above and substituted and got 0=y+(2.2)(2.5)-.5(9.8)(2.5^2) and got 25 for y, but my online homework says it's wrong. I've double checked and can't figure out where I went wrong.
 
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The diver has an initial speed of 2.2m/s in the horizontal direction. What is the diver's initial speed in the vertical direction?
 
zero, because they are not moving either up or down yet, thank you very much!
 

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