If there were no air resistance, with what
speed would drops hit the Earth if they fell
from a cloud 1980 m above the Earth’s sur-
face? The acceleration of gravity is 10 m/s2
v = Vi + a t
Vi = 0
a = -10
so v = -10 t
0 = 1980 + 0 t - (1/2)(10)t^2
5 t^2 = 1980
t = 19.9 seconds
so v = -199 m/s
To find the speed at which drops would hit the Earth if there were no air resistance, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
- v is the final velocity (unknown)
- u is the initial velocity (0 for drops falling from rest)
- a is the acceleration due to gravity (10 m/s^2)
- s is the distance (1980 m)
Plugging the values into the equation, we get:
v^2 = 0 + 2(10)(1980)
v^2 = 39600
Taking the square root of both sides:
v = √39600
v ≈ 198.99 m/s
Therefore, if there were no air resistance, drops falling from a cloud 1980 m above the Earth's surface would hit the Earth with a speed of approximately 198.99 m/s.
To calculate the speed at which the drops would hit the Earth, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v is the final velocity (which is what we want to find)
u is the initial velocity (in this case, 0 m/s since the drop is falling from rest)
a is the acceleration due to gravity (-10 m/s^2 since the drop is accelerating downward)
s is the distance traveled by the drop (1980 m in this case)
Rearranging the equation, we have:
v^2 = 0^2 + 2(-10 m/s^2)(1980 m)
v^2 = 0 + (-39600 m^2/s^2)
v^2 = -39600 m^2/s^2
Since we cannot have a negative value for the velocity, it means there is an error in our calculation. The negative value for the velocity squared tells us that the drops will never hit the Earth if there is no air resistance.
In reality, air resistance plays a significant role in the motion of falling objects, slowing them down as they fall. Without considering air resistance, the drops would continue to accelerate indefinitely, but in the real world, they would reach a terminal velocity when the force of air resistance equals the force of gravity.