A ball is thrown vertically upwards with an initial velocity $v_0$ in a medium where the drag force is proportional to the velocity, i.e., $F_d = -bv$, where b is a positive consta... A ball is thrown vertically upwards with an initial velocity $v_0$ in a medium where the drag force is proportional to the velocity, i.e., $F_d = -bv$, where b is a positive constant. Determine the time it takes for the ball to reach its maximum height. Read more

Steps to Solve

1. Define the forces acting on the ball

The forces acting on the ball are gravity and air resistance (drag force). Gravity acts downwards with a force of $mg$, where $m$ is the mass of the ball and $g$ is the acceleration due to gravity. The drag force acts in the opposite direction to the velocity, so it acts downwards when the ball is going up, and is proportional to the velocity, given by $bv$, where $b$ is a constant.

2. Apply Newton's Second Law

Newton's Second Law states that the sum of the forces acting on an object is equal to its mass times its acceleration ($F = ma$). In this case, let's define upward direction as positive.

$$ma = -mg - bv$$ where $a$ is the acceleration and $v$ is the velocity of the ball.

3. Express acceleration as the derivative of velocity

We can express acceleration as the derivative of velocity with respect to time, $a = \frac{dv}{dt}$. Substituting this into the equation from step 2:

$$m\frac{dv}{dt} = -mg - bv$$

4. Separate variables and integrate

Separate the variables to get $v$ on one side and $t$ on the other:

$$\frac{dv}{mg + bv} = -\frac{dt}{m}$$ Now, integrate both sides: $$\int \frac{dv}{mg + bv} = \int -\frac{dt}{m}$$ The left integral evaluates to $\frac{1}{b} \ln(mg + bv)$, and the right integral evaluates to $-\frac{t}{m} + C$, where C is the integration constant. $$\frac{1}{b} \ln(mg + bv) = -\frac{t}{m} + C$$

5. Solve for the integration constant C

At $t = 0$, the initial velocity is $v_0$. Plug these values into the equation above:

$$\frac{1}{b} \ln(mg + bv_0) = -\frac{0}{m} + C$$ So, $C = \frac{1}{b} \ln(mg + bv_0)$.

6. Substitute C back into the equation

Substituting C back into the equation from step 4:

$$\frac{1}{b} \ln(mg + bv) = -\frac{t}{m} + \frac{1}{b} \ln(mg + bv_0)$$

7. Isolate t

Multiply both sides by $b$: $$\ln(mg + bv) = -\frac{bt}{m} + \ln(mg + bv_0)$$ Rearrange the equation to solve for $t$: $$\frac{bt}{m} = \ln(mg + bv_0) - \ln(mg + bv)$$ $$\frac{bt}{m} = \ln\left(\frac{mg + bv_0}{mg + bv}\right)$$ $$t = \frac{m}{b} \ln\left(\frac{mg + bv_0}{mg + bv}\right)$$

8. Find the time to reach maximum height

At the maximum height, the velocity $v = 0$. Substitute $v = 0$ into the equation from step 7:

$$t = \frac{m}{b} \ln\left(\frac{mg + bv_0}{mg + b(0)}\right)$$ $$t = \frac{m}{b} \ln\left(\frac{mg + bv_0}{mg}\right)$$ $$t = \frac{m}{b} \ln\left(1 + \frac{bv_0}{mg}\right)$$