Solution for Question 16
The air resistance is proportional to velocity: $\vec{a} = -k \vec{v}$.
Consider the horizontal motion from A to C. The horizontal acceleration is:
$$ a_x = \frac{dv_x}{dt} = -k v_x $$
Integrating this gives velocity decay: $v_x = u_x e^{-kt}$.
The horizontal position $x$ is found by integrating velocity:
$$ x(t) = \int_0^t u_x e^{-kt} \, dt = \frac{u_x}{k} (1 – e^{-kt}) $$
The problem states the stone hits C moving almost vertically. This implies the horizontal velocity has decayed to zero, which mathematically corresponds to $t \to \infty$.
At this limit, the total horizontal distance (Range $R$) is:
$$ R = \frac{u_x}{k} (1 – 0) \implies k = \frac{u_x}{R} $$
We analyze the motion above the level line AB. Let $u_y$ be the initial upward velocity at A, and $v_y$ be the downward velocity magnitude at B.
Let $t_1$ be the ascent time (A $\to$ Max Height) and $t_2$ be the descent time (Max Height $\to$ B).
Using Newton’s Second Law and integrating:
$\int_0^{t_1} (g + kv) dt = u_y$
$g t_1 + k \int v dt = u_y$
Since $\int v dt = H$ (Total Vertical Displacement):
$$ g t_1 + kH = u_y \quad \text{— (Eq 1)} $$
DOWNWARD MOTION EQUATION: Force: $mg – mkv$ (Gravity aids, Drag opposes).
$\int_0^{t_2} (g – kv) dt = v_y$
$g t_2 – k \int v dt = v_y$
$$ g t_2 – kH = v_y \quad \text{— (Eq 2)} $$
We are given the differences in velocity and time:
- $\Delta v_y = u_y – v_y$ (Since energy is lost to drag, launch speed > return speed)
- $\Delta t = t_2 – t_1$ (Descent takes longer than ascent due to drag)
Rearranging the equation to solve for $H$: $$ 2kH = \Delta v_y + g \Delta t $$ $$ H = \frac{\Delta v_y + g \Delta t}{2k} $$
Substitute the drag constant $k = \frac{u_x}{R}$ derived in Step 1: $$ H = \frac{\Delta v_y + g \Delta t}{2 (u_x / R)} $$