
#1
Jan2813, 12:34 AM

P: 12

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 Y_{i}=2.00m V_{i}=? a=9.81m/s^2 t= 0.154s Y_{f}= 0m V_{f}=0m/s I 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 



#2
Jan2813, 03:30 AM

Thanks
P: 5,496

The equation you used is good for an unaccelerated motion. The rock's is accelerated. What equations should you use for it?




#3
Jan2813, 12:37 PM

P: 12

could i use the
2a(yy_{i})=V_{y}^{2}V_{oy}^{2}?? 



#4
Jan2813, 12:53 PM

Thanks
P: 5,496

The Rock and the Window
This equation alone will not do it. What is the basic equation of uniformly accelerated motion?



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