How Fast Does a Rock Accelerate Past a Window?

[mgerman63016]

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/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

 

[voko]

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

 

[mgerman63016]

could i use the

2a(y-y_i)=V_y²-V_oy²??

 

[voko]

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

 

[Nickie]

Hi there, as a scientist, I would like to clarify a few things about the scenario you have described. Firstly, it is important to define the units for the measurements you have given. For example, is the initial height of the rock (Yi) measured in meters or feet? This will affect your calculations.

Also, the acceleration due to gravity (a) should be negative because it is acting against the direction of motion of the rock. So, a = -9.81 m/s^2.

Now, let's move on to your equations. The equation you have used, V = Δx/Δt, is correct. However, it is important to note that this equation gives the average velocity of the rock during the time it takes to pass the window. To find the velocity at a specific point, we need to use the equation Vf = Vi + at, where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and t is the time.

To answer the first question, we need to find the final velocity of the rock at the bottom of the window. Since the rock is moving downwards, the final velocity will be negative. So, using the equation Vf = Vi + at, we get Vf = 0 + (-9.81)(0.154) = -1.51 m/s. Therefore, the velocity of the rock at the bottom of the window is -1.51 m/s.

For the second question, we need to find the time it takes for the rock to reach the top of the window. Since we are given the time it takes to pass the window, we can use this time to find the total time it takes for the rock to reach the top of the window. So, the total time (t) will be twice the time it takes to pass the window, which is 2(0.154) = 0.308 seconds.

Finally, to answer the third question, we can use the equation Yi = Vi(t) + 0.5at^2, where Yi is the initial height, Vi is the initial velocity, a is the acceleration, and t is the time. Rearranging this equation, we get Vi = (Yi - 0.5at^2)/t. Plugging in the values we know, we get Vi = (2 - 0.5(-9.81)(0.308