Q: A Particle is projected vertically upward with speed u = 10 m/s . During its journey air applies a force of -0.2 v^{2} on the particle . What is the maximum height attained by the particle [m = 2 kg]

(a) 5 ln2

(b) 3 ln2

(c) 2 ln2

(d) 10 ln2

Ans: (a)

Sol: F = -0.2 v^{2}

$\large m \frac{dv}{dt} = -0.2 v^2 – mg $

$\large \frac{dv}{dt} = -\frac{0.2 v^2}{m} – g $

$\large m \frac{dv}{dt} = -0.1 v^2 – g $ ; (since m = 1 kg)

$\large \int_{10}^{0} \frac{v dv}{0.1 v^2 + g} = – \int_{0}^{y_{max}} dy $

Put , 0.1 v^{2} + g = z

0.1 × 2 v dv = dz

v dv = dz/0.2

When v = 10 , z = 20 and v = 0 , z = 10

$\large \frac{1}{0.2}\int_{20}^{10} \frac{dz}{z} = – \int_{0}^{y_{max}} dy $

$\large \frac{1}{0.2} [ln z]_{20}^{10} = -y_{max} $

$\large \frac{1}{0.2} [ln 10 – ln 20] = -y_{max} $

$\large \frac{1}{0.2} [ln \frac{1}{2}] = -y_{max} $

$\large y_{max} = 5 ln2 $