On a horizontal plane, a projectile of mass \(m\) is launched from the ground with speed \(v_0\) at an angle \(\theta_0\) with the horizontal. In addition to the gravitational force (\(mg\)), it also experiences a drag force \(\vec{F}_{drag} = -\gamma \vec{v}\), where \(\vec{v}\) is its velocity and \(\gamma\) is a constant. It hits the ground at a distance \(R\) from the point of launch with its velocity making an angle \(\theta\) with the horizontal, as shown schematically in the figure below. Then which of the following options is correct?
Show Hint
Solve \(m\dot v_x=-\gamma v_x\) and \(m\dot v_y=-mg-\gamma v_y\) separately. \(v_x\) just decays to zero, \(v_y\) settles at a finite terminal speed, so the landing ratio \(|v_y|/v_x\) grows, and the drag removes energy so the range shrinks.
Step 1: Start from the work-energy theorem.
Compare this projectile with an identical one launched at the same $v_0$ and $\theta_0$ but with no drag. Drag always points opposite to the velocity, so it does negative work on the real projectile at every single instant of flight: $dW_{drag}/dt = -\gamma v^2 \lt 0$. That means the drag-affected projectile always has less kinetic plus potential energy available than its drag-free twin, at every point along the path.
Step 2: Turn the energy loss into a statement about the range.
With less energy available throughout the flight, the drag-affected projectile cannot sustain the same horizontal speed as its drag-free twin at any time after launch (its $v_x$ only ever decays, since drag is the only horizontal force acting). Integrating a strictly smaller $v_x(t)$ over the flight gives a horizontal distance strictly less than the drag-free range:
\[ R = \int_0^{T} v_x(t)\, dt \ \lt \ \int_0^{T_0} v_0\cos\theta_0 \, dt = \frac{v_0^2 \sin 2\theta_0}{g} \]
So $R \lt \dfrac{v_0^2 \sin 2\theta_0}{g}$ regardless of the exact value of $\gamma$.
Step 3: Now look at the landing angle using terminal velocity.
Along the vertical direction, gravity and drag combine so that $v_y$ approaches a fixed terminal value $v_t = mg/\gamma$ (downward) as time goes on, it never grows without bound the way free-fall speed does. Along the horizontal direction there is nothing but drag acting, so $v_x$ has no floor to level off at, it keeps shrinking toward zero the longer the flight lasts.
Step 4: Combine the two trends.
Because $v_x$ keeps shrinking while $v_y$ settles near a fixed nonzero terminal value, the ratio $\tan\theta = |v_y|/v_x$ can only grow as the flight goes on, it is larger at landing than it was at launch, where $\tan\theta_0 = |v_y(0)|/v_x(0)$. So the ball must land steeper than it left: $\theta \gt \theta_0$.
Final Answer:
The energy argument fixes the range, $R \lt v_0^2\sin 2\theta_0/g$, and the terminal-velocity argument fixes the angle, $\theta \gt \theta_0$, together giving option (C).
\[ \boxed{\text{Option (C): } R \lt \frac{v_0^2\sin 2\theta_0}{g}, \ \theta \gt \theta_0} \]
