A flower pot from a roof top garden fall off the edge of the roof and fall fast the window below. Air resistance may be ignored. The flower pot takes 0.48s to travell between the top and the bottom of the window which is 1.9m high. How far below the roof is the top of the window?
To solve this problem, we can use the equations of motion. We know that the flower pot falls freely under the influence of gravity, so we can use the equation:
y = 0.5 * g * t^2
Where:
y is the vertical distance traveled (1.9m in this case),
g is the acceleration due to gravity (approximately 9.8 m/s^2), and
t is the time taken to travel the distance (0.48s in this case).
Rearranging the equation to solve for t, we get:
t = sqrt(2y / g)
Substituting the values we know:
t = sqrt(2 * 1.9 / 9.8)
t = sqrt(0.387755102)
t ≈ 0.622s
Now, to find the distance between the roof and the top of the window, we need the average velocity. We can use the equation:
v_avg = y / t
Substituting the values we know:
v_avg = 1.9 / 0.622
v_avg ≈ 3.05 m/s
Since we are ignoring air resistance, the acceleration is constant. Therefore, the average velocity is equal to half of the final velocity. So, we can use the equation:
v_avg = (v_initial + v_final) / 2
Since the flower pot starts from rest, the initial velocity (v_initial) is 0. Therefore:
3.05 = v_final / 2
Solving for v_final:
v_final = 3.05 * 2
v_final ≈ 6.1 m/s
Now, we can use the equation of motion:
v_final = v_initial + g * t
Since the final velocity (v_final) is directed downward, it is negative:
-6.1 = 0 + (-9.8) * t
Solving for t:
t = -6.1 / (-9.8)
t ≈ 0.622s
Comparing this with the given time of 0.48s, we can conclude that our initial assumption of the flower pot falling for 0.48s was incorrect. The correct time is approximately 0.622 seconds.
Therefore, the distance below the roof is not provided in the information given.
As the ball falls, the distance fallen at time t is 4.9t^2
Now, we know that at some time t, it falls past the top of the window, and then .048 seconds later, it has fallen 1.9m more. So,
4.9t^2 + 1.9 = 4.9(t+0.48)^2
t = 0.1639
That is, it fell for 0.1639 seconds before passing the top of the window.
So, how far did the ball fall during the first 0.1639 seconds?