If there were no air resistance, with what
speed would drops hit the Earth if they fell
from a cloud 3881 m above the Earth’s surface?
The acceleration of gravity is 10 m/s
3881 = 4.9 t^2
t = 28.1 seconds
v = g t = 9.81 (28.1)
= 276 m/s
To find the speed at which drops would hit the Earth with no air resistance, we can use the equations of motion and the concept of free fall.
First, we need to determine the time it takes for the drops to fall from the cloud to the Earth's surface. We can use the equation:
s = ut + (1/2)gt^2
Where:
s = distance (3881 m in this case)
u = initial velocity (0 m/s as the drops start from rest)
g = acceleration due to gravity (10 m/s^2)
t = time (unknown)
Rearranging the equation to solve for time (t), we get:
t = √(2s/g)
Plugging in the given values:
t = √(2 * 3881 m / 10 m/s^2)
t = √(7762 m / 10 m/s^2)
t = √(776.2 s^2)
t ≈ 27.88 seconds
Now that we have the time taken, we can determine the final velocity (v) using the equation:
v = u + gt
Plugging in the values:
v = 0 m/s + 10 m/s^2 * 27.88 s
v = 0 m/s + 278.8 m/s
v ≈ 278.8 m/s
Therefore, if there were no air resistance, drops would hit the Earth with a speed of approximately 278.8 m/s.