Fall time under g is sqrt 2h/g. Can someone give a derivation from the equations of motion please? Thanks.
Sure! I can explain how to derive the formula for the fall time under gravity using the equations of motion.
The equation of motion that we will use is the one that relates the distance traveled (h), the initial velocity (u), the acceleration due to gravity (g), and the time taken (t):
h = ut + (1/2)gt^2
In this case, we assume that the object starts from rest (u = 0), so the equation simplifies to:
h = (1/2)gt^2
To find the fall time, we need to solve for t. Rearranging the equation, we have:
2h = gt^2
Now, let's isolate t by dividing both sides of the equation by g:
2h/g = t^2
To find t, we need to take the square root of both sides of the equation:
t = sqrt(2h/g)
And there you have it! The derived equation for the fall time (t) in terms of the distance fallen (h) and the acceleration due to gravity (g) is:
t = sqrt(2h/g)
This formula tells us that the fall time is proportional to the square root of the distance fallen, and inversely proportional to the square root of the acceleration due to gravity.
Sure! I can provide you with a step-by-step derivation of the equation for fall time under the acceleration due to gravity.
Let's start with the equations of motion for an object in free fall:
1) v = u + gt (equation 1)
2) s = ut + 0.5gt^2 (equation 2)
Where:
v = final velocity
u = initial velocity
g = acceleration due to gravity
t = time
s = displacement
In the case of an object falling from rest, the initial velocity (u) is equal to 0. Therefore, equation 1 becomes:
v = gt (equation 3)
Now, let's consider the situation where the object falls from a certain height 'h'. At the top of its path, the displacement (s) is 0. Therefore, equation 2 becomes:
0 = 0 + 0.5gt^2 (equation 4)
Now, solve equation 4 for t:
0.5gt^2 = 0
t^2 = 0
Since time cannot be negative, we can ignore the solution t = 0. Therefore, t = √0 = 0.
This means that the object takes 0 seconds to reach the top of the path.
Next, let's consider the time it takes for the object to fall from the top to the bottom of the path. At the bottom of the path, the displacement (s) is equal to h (the height). Substituting this into equation 2, we get:
h = 0 + 0.5gt^2
2h = gt^2
t^2 = 2h/g
t = √(2h/g)
Therefore, the derivation shows that the fall time (t) under the acceleration due to gravity (g) can be represented by:
t = √(2h/g)
This is the equation you're looking for!
I hope this explanation was helpful. Let me know if there's anything else I can assist you with!