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: Understanding the Concept:
Without any drag, a projectile launched from level ground follows a symmetric parabola. Its horizontal range is the standard formula \(R_0 = \dfrac{v_0^2 \sin 2\theta_0}{g}\), and by the up-down symmetry of that parabola the ball lands at exactly the same angle it was launched at, so \(\theta = \theta_0\). Here the projectile also feels a drag force \(\vec{F}_{drag} = -\gamma \vec{v}\), proportional to its own velocity, so this symmetry is broken and we have to track the horizontal and vertical motion separately.
Step 2: Key Formula or Approach:
Newton's second law along each axis, with drag as the only horizontal force and gravity plus drag along the vertical, gives two independent linear equations:
\[ m\frac{dv_x}{dt} = -\gamma v_x, \qquad m\frac{dv_y}{dt} = -mg - \gamma v_y \]
Solving the first with \(v_x(0) = v_0\cos\theta_0\) gives an exponential decay, and solving the second (a linear equation with a constant forcing term \(-mg\)) gives a term that settles down to a constant terminal value:
\[ v_x(t) = v_0\cos\theta_0\, e^{-\gamma t/m}, \qquad v_y(t) = \left(v_0\sin\theta_0 + \frac{mg}{\gamma}\right) e^{-\gamma t/m} - \frac{mg}{\gamma} \]
Step 3: Why the landing angle is steeper (\(\theta \gt \theta_0\)):
Look at what each velocity component does as time goes on. \(v_x(t)\) has nothing driving it except drag, so it just keeps decaying toward zero, no matter how long the flight lasts. \(v_y(t)\), on the other hand, is pulled by gravity the whole time and only levels off at the terminal speed \(-mg/\gamma\) as \(t \to \infty\); it does not collapse toward zero the way \(v_x\) does. So while both speeds shrink compared with the no-drag case, the horizontal speed shrinks away much faster (all the way to zero) than the vertical speed does (it only settles at a finite terminal value). At the moment of landing, the ratio \(\tan\theta = |v_y|/v_x\) has therefore grown larger than it was at launch, \(\tan\theta_0 = |v_y(0)|/v_x(0)\). That means \(\theta \gt \theta_0\): the ball comes down steeper than it went up.
Step 4: Why the range is shorter (\(R \lt v_0^2\sin 2\theta_0/g\)):
Drag is a dissipative force: it removes mechanical energy from the projectile at the rate \(\gamma v^2\) at every instant of the flight, energy that in the no-drag case would have stayed available to carry the projectile further forward. Since \(v_x(t)\) is smaller at every instant than the drag-free value \(v_0\cos\theta_0\), the horizontal distance \(R = \int_0^T v_x\,dt\) covered in the (also reduced) flight time \(T\) comes out smaller than the drag-free range for the same launch speed and angle. So \(R \lt \dfrac{v_0^2\sin 2\theta_0}{g}\).
Final Answer:
Both effects act together: the range shrinks below the ideal value and the ball lands steeper than it was thrown, which is exactly option (C).
\[ \boxed{R \lt \frac{v_0^2\sin 2\theta_0}{g}, \ \ \theta \gt \theta_0 \ \ \text{(Option C)}} \]
