NLM BYU 45

Solution 45

Let the mass of the ball be $m$. The forces acting on the ball are gravity and air resistance.

Given: Air resistance is proportional to speed. Let $\vec{F}_{air} = -k\vec{v}$, where $k$ is a constant and $\vec{v}$ is the velocity vector.

Using Newton’s Second Law:

$$ \vec{F}_{net} = m\vec{g} + \vec{F}_{air} $$ $$ m\frac{d\vec{v}}{dt} = m\vec{g} – k\vec{v} $$

We integrate this equation with respect to time over the total time of flight $T$. Let the initial velocity be $\vec{u}$ (upwards) and the final velocity be $\vec{v}_{f}$ (downwards).

$$ \int_{t=0}^{t=T} m \, d\vec{v} = \int_{0}^{T} m\vec{g} \, dt – \int_{0}^{T} k\vec{v} \, dt $$ $$ m(\vec{v}_{f} – \vec{u}) = m\vec{g} T – k \int_{0}^{T} \vec{v} \, dt $$

The term $\int_{0}^{T} \vec{v} \, dt$ represents the total displacement of the ball. Since the ball returns to the ground, the net vertical displacement is zero.

$$ \int_{0}^{T} \vec{v} \, dt = 0 $$

Thus, the equation simplifies to:

$$ m(\vec{v}_{f} – \vec{u}) = m\vec{g} T $$ $$ \vec{v}_{f} – \vec{u} = \vec{g} T $$

Let’s define the upward direction as positive ($+\hat{j}$).

• Initial velocity $\vec{u} = +20 \, \hat{j}$ m/s
• Final velocity $\vec{v}_{f} = -16 \, \hat{j}$ m/s (downward)
• Acceleration due to gravity $\vec{g} = -10 \, \hat{j}$ m/s²

Substituting these values:

$$ (-16) – (20) = (-10) T $$ $$ -36 = -10 T $$ $$ T = 3.6 \, \text{s} $$