using d2h/dt2=−g+k(�dh/dt)^2, find an expression for the terminal velocity in terms of k and g?
To find an expression for the terminal velocity in terms of k and g, we need to solve the given differential equation:
d²h/dt² = -g + k(dh/dt)²
Let's break down the steps to find the expression:
Step 1: Separate the variables
Rearrange the equation to isolate the variables on each side:
(dh/dt)² = (1/k)(d²h/dt² + g)
Step 2: Take the square root
Take the square root of both sides to eliminate the square:
dh/dt = ±√[(1/k)(d²h/dt² + g)]
Step 3: Integrate
Integrate both sides with respect to time (t):
∫(1/√[(1/k)(d²h/dt² + g)]) dh = ∫dt
This integration will require some algebraic manipulation and substitutions.
Step 4: Evaluate the integral
To evaluate the integral, let's assume that h = f(t), so we can use the chain rule.
Let y = d²h/dt² + g. Then dy/dt = d³h/dt³.
Rewriting the equation, we have (dh/dt)³ = k/y.
Now, let u = dh/dt, we can write the equation as u²du = k/y dt.
Integration will give us: (1/3)u³ = k ln|y| + c₁, where c₁ is the constant of integration.
Step 5: Substitute back
Substitute y = d²h/dt² + g and u = dh/dt back into the equation:
(1/3)(dh/dt)³ = k ln|(d²h/dt² + g)| + c₁
Step 6: Find the terminal velocity
At terminal velocity, the object is no longer accelerating, so d²h/dt² = 0. Plug this into the equation:
(1/3)(dh/dt)³ = k ln|g| + c₁
Since the object is at terminal velocity, dh/dt = 0. Substituting it into the equation, we get:
k ln|g| + c₁ = 0
c₁ = -k ln|g|
Now, substitute c₁ back into the equation:
(1/3)(dh/dt)³ = k ln|g| - k ln|g|
(1/3)(dh/dt)³ = 0
dh/dt = 0
Therefore, the expression for the terminal velocity in terms of k and g is dh/dt = 0.
To find the expression for the terminal velocity, we need to first understand the concept of terminal velocity. Terminal velocity is the maximum velocity that an object can reach when the drag force acting on it is equal to the gravitational force pulling it down.
In this case, the equation provided d2h/dt2 = -g + k(dh/dt)^2 represents the second-order differential equation for the height of the object (h) with respect to time (t). The negative sign on the left side (-g) represents the force of gravity pulling the object downward, and the positive term on the right side (k(dh/dt)^2) represents the drag force opposing the motion.
To determine the expression for the terminal velocity, we need to find the point at which the object stops accelerating and reaches a constant velocity. This can be done by setting d2h/dt2 = 0 and solving for dh/dt.
Setting d2h/dt2 = 0, we have:
0 = -g + k(dh/dt)^2
Rearranging the equation, we get:
g = k(dh/dt)^2
Taking the square root of both sides, we have:
√g = √k(dh/dt)
Simplifying further, we get:
√k(dh/dt) = √g
Now, we can integrate both sides:
∫(1/√k) dh = ∫√g dt
Integrating both sides yields:
(2/√k)h + C1 = 2√g t + C2
Where C1 and C2 are constants of integration.
At terminal velocity, dh/dt = 0, which means the object is not changing its height with time. Therefore, substituting dh/dt = 0 into our equation, we get:
(2/√k)h + C1 = 0 + C2
Simplifying further, we have:
(2/√k)h = C2 - C1
Let's define another constant C3 = C2 - C1, which gives us:
(2/√k)h = C3
Rearranging the equation, we finally obtain the expression for the terminal velocity (v):
v = h/t = C3 * √k/2
So, the expression for the terminal velocity in terms of k and g is:
v = C3 * √k/2