How long does it take for a stone thrown from a building to pass a window?

[coglon]

trying to study for a midterm with kinematics 1d 2d and dynamics
Heres problem:

A stone is thrown vertically downward from the top of a 96m building at a velocity of 5.0m/s. This stone is observed passing by a window that has a height of 2.0m. Assuming that the bottom of the window is 25m from the ground, how long (time) does it take for the stone to pass the window?

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Heres what I've done so far:

distance = d = 96m
time= t = ?
constant velocity = v = 5.0m/s
acceleration = a = 9.81m/s^2
so i used v=d/t to find total time (5)=(96)/(t) I got 19.2 seconds.
So to find the time it takes to pass the window i was thinking of using a kinematics equation. Because we know the window is 25m from the ground we know that during the last 25m of free fall that velocity initial = vi = 5.0m/s and velocity final = vf = 0 m/s
OK so i think the above time is wrong and using a kinematics equation i found a different total using gravity, distance etc.. I got 4.424 seconds. well I am sort of lost... anyways I was pugging in a bunch of numbers into different formula.. I found the bottom 25m and the 69m above the 2 m window and tried a subtraction... but got wrong answer... it was negative... well could some one show me how to do this question the right way..

BTW.. the real answer is: (5.3x10^-2 sec)

Thanks for help

 

[PrudensOptimus]

Δy = v₀t - 0.5gt^2

-69 = 5t - 0.5gt^2

-13.8 = t - t^2

t = 4.248332962798262403050998906414 when the object reaches the top of the window.

 

[coglon]

"Äy = v0t - 0.5gt^2

-69 = 5t - 0.5gt^2

-192 = 10t - 10t^2

t^2 - t - 19.2 = 0

t = 4.9 when the object reaches the top of the window." - you wrote


so i really don't get your equation here... how does 0.5gt^2 give you 10t^2... even if you used 10 for gravity you would still only get 5t^2 and even if your right how do you get 4.9 from "t^2 - t - 19.2=0" that won't work because you got 2 unknown variables...or something like that the closest i could get was 4.4 s

 

[PrudensOptimus]

It's called a 2nd degree equation, you can solve it by applying the quadratic formula. Equation of the 2nd degree is not a multivariable equation, it is solvable.

It is often very useful to qualify the mathematics needed for physics, rather than learning them at the same time.

 

[Doc Al]

Originally posted by PrudensOptimus
Δy = v₀t - 0.5gt^2

-69 = 5t - 0.5gt^2

-13.8 = t - t^2

t = 4.248332962798262403050998906414 when the object reaches the top of the window.

Take care to use a consistent sign convention. Choosing the origin to be at ground level, and up to be positive, I get:
y = 96 -5t - 0.5gt^2 (note: y₀= 96)

at the top of the window: y = 27, so:

27 = 96 -5t - .5(9.8)t^2

Solving the quadratic for t, I get t = 3.277 secs

 

[Meab5]

A stone is thrown vertically downward from the top of a 96m building at a velocity of 5.0m/s. This stone is observed passing by a window that has a height of 2.0m. Assuming that the bottom of the window is 25m from the ground, how long (time) does it take for the stone to pass the window?


for this problem i would find the velocity of the stone at the top of the window and then the velocity at the bottom of the window and the get the average and divide the size of the window by the average velocity.

V^2=(Vnaught)^2 + 2G(Y-Ynaught) V^2=25 + 2(-9.8)(69)=34.43 m/s this is the velocity at the top of the window

using the same equation you can find that at the bottom of the window it is moving at 36.97 m/s

the average velocity is 35.72

using v=x/t where x is displacement you can move the variables around to get t=x/v t=2/35.72 t=.056 s which if i hadn't rounded the velocities only to two digits would have equalled the .053s in the answer book

Hope this helps

 

[jamesrc]

Meab5: Your method is valid, but you need to heed Doc Al's previous warning about sign conventions. If you're going to use a = -9.8 m/s/s, then in your equation y = 27 (or 25, depending on whether your looking at the top or the bottom of the window) and yo = 96 (not the other way around as you had it). This will affect your calculated velocities a bit: vtop = 37.13 m/s and vbot = 37.65 m/s. You can use these to find the total time as you suggested, or you could find the times at which the two events occur and find the difference (as previously suggested); same difference.

Oh, and PrudensOptimus: I take it you're not an engineer ? I've been known to play fast and loose with the rules for significant digits myself, but thirty-some odd digits is a bit much .

 

[PrudensOptimus]

Originally posted by jamesrc

Oh, and PrudensOptimus: I take it you're not an engineer ? I've been known to play fast and loose with the rules for significant digits myself, but thirty-some odd digits is a bit much .

I try my best... I am a sophmore in high school.