How Fast Does a Rock Accelerate Past a Window?

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A rock dropped from a building takes 0.154 seconds to fall past a window measuring 2.00 meters tall. The velocity of the rock at the bottom of the window is calculated to be 12.98 m/s, but the initial estimation was incorrect. To accurately determine the time taken to reach the top of the window and the height from which it was dropped, the correct equations for uniformly accelerated motion must be applied. The discussion emphasizes the need to consider acceleration due to gravity, which is 9.81 m/s². Proper guidance on using the right equations for accelerated motion is sought for accurate calculations.
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1. A rock is dropped off the top of a building. On the way down, the rock passes a window. The window is known to be tall and the stone takes 0.154 to fall past the window.

1. What is the velocity of the rock at the bottom of the window?
2. How much time was necessary from the instant the rock was dropped until it reached the top of the window?
3. How far above the top of the window was the rock dropped?


So far I got
Yi=2.00m
Vi=?
a=9.81m/s^2
t= 0.154s
Yf= 0m
Vf=0m/sI also got the initial velocity for the top window to be 12.98m/s which because I got my estimation wrong I got the answer wrong...sigh...BUT if I can get a little guidance I would GREATLY appreciate it! THANKS

And for my equation I used V = Δx/Δt = 2.00/0.154 = 12.98 m/s
 
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The equation you used is good for an unaccelerated motion. The rock's is accelerated. What equations should you use for it?
 
could i use the

2a(y-yi)=Vy2-Voy2??
 
This equation alone will not do it. What is the basic equation of uniformly accelerated motion?
 

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