Question

A projectile of mass 200 g is launched in a viscous medium at an angle $60^\circ$ with the horizontal, with an initial velocity of 270 m/s. It experiences a viscous drag force $\vec{F} = -c \vec{v}$ where the drag coefficient $c = 0.1 \, \text{kg/s}$ and $\vec{v}$ is the instantaneous velocity of the projectile. The projectile hits a vertical wall after 2 s. Taking $e = 2.7$, the horizontal distance of the wall from the point of projection (in m) is ____

Hint:

In viscous drag proportional to velocity, horizontal velocity decays exponentially and displacement is given by integrating velocity over time.

Answer 1

Step 1: The motion is affected by viscous drag proportional to velocity, so horizontal velocity decreases as: \[ m \frac{dv_x}{dt} = - c v_x \implies \frac{dv_x}{dt} = -\frac{c}{m} v_x \]
Step 2: Solve this first-order differential equation: \[ v_x(t) = v_{x0} e^{-\frac{c}{m} t} \] where initial horizontal velocity is \[ v_{x0} = v_0 \cos 60^\circ = 270 \times \frac{1}{2} = 135 \text{ m/s} \]
Step 3: Horizontal displacement in time \(t\) is \[ x(t) = \int_0^t v_x(t) dt = \int_0^t v_{x0} e^{-\frac{c}{m} t} dt = v_{x0} \frac{m}{c} (1 - e^{-\frac{c}{m} t}) \]
Step 4: Plugging in values: \[ m = 0.2 \text{ kg}, \quad c = 0.1 \text{ kg/s}, \quad t=2 \text{ s}, \quad e=2.7 \] Calculate \[ \frac{c}{m} = \frac{0.1}{0.2} = 0.5 \] Calculate \[ e^{-\frac{c}{m} t} = e^{-0.5 \times 2} = e^{-1} = \frac{1}{e} = \frac{1}{2.7} \] Step 5: Calculate horizontal distance: \[ x = 135 \times \frac{0.2}{0.1} \times \left(1 - \frac{1}{2.7}\right) = 135 \times 2 \times \left(1 - 0.3704\right) = 270 \times 0.6296 = 170 \text{ m (approx.)} \] Step 6: The projectile hits the vertical wall at approximately 170 m from the point of projection.