First slide

Vertical projection from top of a tower

Question

A particle is projected vertically upward with speed u = 10 m/s. During its journey air applies a force of -0.2v² on the particle. What is the maximum height attained by the particle [m = 2kg]

Moderate

A

5 ln (2)
B

3ln (2)
C

2ln (2)
D

10 ln (2)

Solution

$a = - v \frac{dv}{dy} = - g - \frac{0 .2 v^{2}}{2}$
$\Rightarrow dy = - \frac{10 v dv}{100 + v^{2}}$
$\Rightarrow y_{\max} = - 5 \int_{10}^{0} \frac{2 v dv}{100 + v^{2}}$
$\Rightarrow y_{\max} = - 5 {[\ln [100 + v^{2}]]}_{10}^{0}$
= 5 ln [2]

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