A stone is dropped from rest at the top of a tall building. Determine the distance the ball has traveled after ½ second, one second, 3/2 seconds, and 2 seconds. How fast is the ball moving at the end of the 2 seconds?
d = (1/2) g t^2 = 4.9 t^2
if t = 1/2 for example
d = 4.9/4 = 1.225 meters
to get speed
v = g t = 9.8 t
if t = 2 seconds
v = 19.6 meters/second
To determine the distance traveled by the stone at different times, we need to use the equations of motion under constant acceleration. In this case, the stone is dropped from rest, so its initial velocity (u) is 0, and the only force acting on it is gravity.
The equation that relates distance (s), initial velocity (u), time (t), and acceleration (a) is:
s = ut + (1/2)at²
In this case, the acceleration due to gravity (a) is approximately 9.8 m/s² (assuming no air resistance).
Let's calculate the distance traveled by the stone at different times:
1. After ½ second (t = 0.5 s):
s = (0) * (0.5) + (1/2) * (9.8) * (0.5)²
s = 0 + (1/2) * (9.8) * (0.25)
s = (1/2) * (9.8) * (0.25)
s ≈ 1.225 meters
2. After one second (t = 1 s):
s = (0) * (1) + (1/2) * (9.8) * (1)²
s = 0 + (1/2) * (9.8) * (1)
s = (1/2) * (9.8)
s ≈ 4.9 meters
3. After 3/2 seconds (t = 1.5 s):
s = (0) * (1.5) + (1/2) * (9.8) * (1.5)²
s = 0 + (1/2) * (9.8) * (2.25)
s = (1/2) * (9.8) * (2.25)
s ≈ 10.0125 meters
4. After 2 seconds (t = 2 s):
s = (0) * (2) + (1/2) * (9.8) * (2)²
s = 0 + (1/2) * (9.8) * (4)
s = (1/2) * (9.8) * (4)
s ≈ 19.6 meters
Now, let's determine the speed of the stone at the end of 2 seconds. The final speed (v) can be found using the formula:
v = u + at
Here, the initial velocity (u) is 0, and the acceleration (a) is still 9.8 m/s². Plugging in these values:
v = 0 + (9.8) * (2)
v = 0 + 19.6
v ≈ 19.6 m/s
Therefore, the stone is moving at a speed of approximately 19.6 m/s at the end of 2 seconds.