Question 15

NUMERICALMEDIUM

A projectile of mass 200200 g is launched in a viscous medium at an angle 60∘60^\circ with the horizontal, with an initial velocity of 270270 m/s. It experiences a viscous drag force F⃗=−cv⃗\vec{F} = -c\vec{v} where the drag coefficient c=0.1c = 0.1 kg/s and v⃗\vec{v} is the instantaneous velocity of the projectile. The projectile hits a vertical wall after 22 s. Taking e=2.7e = 2.7, the horizontal distance of the wall from the point of projection (in m) is ____

Correct Answer: 170

Detailed Solution

  1. Parameters: m=0.2m = 0.2 kg, c=0.1c = 0.1 kg/s, v0=270v_0 = 270 m/s, θ=60∘\theta = 60^\circ, t=2t = 2 s.
  2. Initial horizontal velocity: vx0=v0cos⁡60∘=270×0.5=135v_{x0} = v_0 \cos 60^\circ = 270 \times 0.5 = 135 m/s.
  3. Horizontal equation of motion: mdvxdt=−cvx  ⟹  vx(t)=vx0e−cmtm \frac{dv_x}{dt} = -c v_x \implies v_x(t) = v_{x0} e^{-\frac{c}{m}t}.
  4. Substitute constants: cm=0.10.2=0.5\frac{c}{m} = \frac{0.1}{0.2} = 0.5 s−1^{-1}. So, vx(t)=135e−0.5tv_x(t) = 135 e^{-0.5t}.
  5. Horizontal distance x=∫02135e−0.5tdt=135[e−0.5t−0.5]02=270(1−e−1)x = \int_0^2 135 e^{-0.5t} dt = 135 \left[ \frac{e^{-0.5t}}{-0.5} \right]_0^2 = 270 (1 - e^{-1}).
  6. Using e=2.7e = 2.7: x=270(1−12.7)=270(1.72.7)=100×1.7=170x = 270 (1 - \frac{1}{2.7}) = 270 (\frac{1.7}{2.7}) = 100 \times 1.7 = 170 m.
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