A particle is projected vertically upward with u = 10text m/s . Air exerts a resistive force F = -0.2v^2 on it (m = 2text kg , g = 10text m/s ^2). The maximum height attained is:

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NEET Physics QUESTION #7324

Question 1

A particle is projected vertically upward with $u = 10\text{ m/s}$. Air exerts a resistive force $F = -0.2v^2$ on it ($m = 2\text{ kg}$, $g = 10\text{ m/s}^2$). The maximum height attained is:

$5\ln 2$✔️
$3\ln 2$
$2\ln 2$
$10\ln 2$

Correct Answer Explanation

Net force (taking up as positive): $F_{net} = -mg - 0.2v^2 = -20 - 0.2v^2$

Using $ma = v\dfrac{dv}{dx}$: $2v\dfrac{dv}{dx} = -20 - 0.2v^2$

Separating variables: $\dfrac{v\,dv}{20+0.2v^2} = -\dfrac{dx}{2}$

Integrating from $v = 10$ to $v = 0$: let $k = 20 + 0.2v^2$, $dk = 0.4v\,dv$

$\dfrac{1}{0.4}\big[\ln(20) - \ln(40)\big] = -\dfrac{H}{2} \Rightarrow H = \dfrac{2\ln 2}{0.4} = \mathbf{5\ln 2}$

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