a 2kg mass is dropped from 12km height at what height would the velocity be 5m/s?
I assume you mean 12 meters, not 12 kilometers
in general for constant acceleration a:
(the mass has nothing to do with this question)
v = Vi + a t
h = Hi + Vi t + (1/2) a t^2
here
v = - g t = -9.81 t
h = 12 - 4.9 t^2
so for v = 5 (actally-5, it goes down)
-5 = -9.81 t
t = 0.51 second
h = 12 - 4.9(0.51)^2=12 - 1.27 = 10.7 meters
To solve this problem, we can use the equations of motion for an object in free fall.
The first equation we can use is the equation for the final velocity of an object in free fall:
v^2 = u^2 + 2as
Where:
v = final velocity (5 m/s in this case)
u = initial velocity (0 m/s, since the object is dropped from rest)
a = acceleration due to gravity (approximately 9.8 m/s^2)
s = distance traveled (the height at which the velocity is 5 m/s, which we will solve for)
Rearranging the equation gives us:
s = (v^2 - u^2) / (2a)
Plugging in the values, we have:
s = (5^2 - 0^2) / (2 * 9.8)
s = 25 / 19.6
s ≈ 1.28 m
Therefore, at a height of approximately 1.28 meters, the velocity of the 2 kg mass would be 5 m/s.
To determine at what height the velocity of the 2 kg mass would be 5 m/s, we can use the principles of physics involving free-fall motion and conservation of energy.
Step 1: Calculate the initial velocity:
Since the mass is dropped from rest, the initial velocity (u) is 0 m/s.
Step 2: Calculate the final velocity:
Given the final velocity (v) as 5 m/s.
Step 3: Calculate the acceleration due to gravity:
The acceleration due to gravity (g) on Earth is approximately 9.8 m/s².
Step 4: Use the equations of motion for free fall:
The equation we can use is v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Re-arranging the equation, we have s = (v² - u²) / (2a)
Substituting the known values:
s = (5² - 0²) / (2 * 9.8)
s = 25 / 19.6
s ≈ 1.28 meters
Thus, the mass would reach a velocity of 5 m/s at a height of approximately 1.28 meters.