A stone is observed falling past a small window 1 meter tall in 0.2 seconds. how fast is it moving as it passes the bottom of the window?how far is the bottom of the window from the roof, where the stone originally began its fall from rest?what is the stone's average speed as it passes the window?
i can't figure out a way to do it, can any one please give me some suggestions????
THX A LOT!!!!
distance = Vot + 1/2(g)t^2
1 m = Vo(0.2) + 1/2(9.8)*(0.2)^2
solve for Vo which is the velocity at the top of the window. I have 4.02 m/s at the top of the window.
Use Vo as V for V = Vo + gt. At rest Vo = 0 and
4.02 + 9.8*t and solve for t, the time to fall from rest to the top of the window.
I have 0.41 seconds.
then distance = Vo*t + 1/2(9.8)(0.41)^2
Vo is 0 at the beginning. Check my thinking. Check my work.
Following your thinking,
I first got Vo=4.02 m/s = the speed at the top of the window
then I use the equation:Vf=Vo+gt,
so I got Vf=5.981m/s= the speed at the bottom of the window
and then I use the equation: Vf^2-Vo^2=2gd,
thus I got d=1.82m = the distance from the roof to the bottom of the window
at last, I use the equation: (Vf+Vo)/2,
finally I got the average speed as the stone passes the window = 5m/s
Is it correct?
My numbers were the same as yours.
To solve these questions, we can use the basic equations of motion and the concepts of displacement, velocity, and acceleration.
1. How fast is the stone moving as it passes the bottom of the window?
To find the speed of the stone as it passes the bottom of the window, we need to determine its final velocity. Since the stone is observed falling past a 1-meter tall window in 0.2 seconds, we can use the equation of motion for constant acceleration:
v = u + at
Where:
v = final velocity
u = initial velocity (in this case, 0 m/s since the stone starts from rest)
a = acceleration due to gravity (approximately 9.8 m/s^2)
t = time (0.2 seconds in this case)
Substituting the known values:
v = 0 + (9.8 m/s^2)(0.2 s)
v = 1.96 m/s
Therefore, the stone is moving with a speed of 1.96 m/s as it passes the bottom of the window.
2. How far is the bottom of the window from the roof, where the stone originally began its fall from rest?
To find the distance from the roof to the bottom of the window, we need to determine the displacement of the stone during its fall. Again, using the equation of motion:
s = ut + 0.5at^2
Where:
s = displacement (unknown in this case)
u = initial velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2 since the stone is falling)
t = time (0.2 seconds)
Substituting the known values:
s = 0(0.2) + 0.5(-9.8)(0.2)^2
s = -0.196 m
Here, the negative sign indicates that the displacement is downward, as expected during free fall. Therefore, the bottom of the window is located approximately 0.196 meters below the roof.
3. What is the stone's average speed as it passes the window?
Average speed is defined as the total distance traveled divided by the total time taken. Since the stone falls past a 1-meter tall window in 0.2 seconds, and we have already determined the distance to be 0.196 meters, we can calculate the average speed:
Average Speed = Total Distance / Total Time
Average Speed = (1.00 m + 0.196 m) / 0.2 s
Average Speed = 1.196 m / 0.2 s
Average Speed = 5.98 m/s
Therefore, the stone's average speed as it passes the window is approximately 5.98 m/s.
By using these equations and concepts, you should be able to solve the given questions.