First slide

Projection Under uniform Acceleration

Question

A particle of mass 1 kg is fired with velocity 50 m/s at an angle of 60° from horizontal. It is acted by viscous force of 0.2v during its journey. The horizontal distance travelled by it in first 10 seconds is :

Difficult

A

90 m
B

108 m
C

125 m
D

213 m

Solution

$\frac{{dv}_{x}}{dt} = - \frac{v}{5} cosθ = - \frac{v_{x}}{5}$
$\Rightarrow \int \frac{{dv}_{x}}{v_{x}} = - \int \frac{1}{5} dt$
$\Rightarrow {({Inv}_{x})}_{u_{x}}^{v_{x}} = {(- \frac{t}{5})}_{0}^{t} \Rightarrow In (\frac{v_{x}}{u_{x}}) = - \frac{t}{5}$
$\Rightarrow v_{x} = u_{x} e^{- t / 5} = (50 \cos 60 °) e^{- t / 5} = 25 e^{- t / 5}$
$Now \frac{dx}{dt} = v_{x} = 25 e^{- t / 5}$
$\Rightarrow x = 25 (\frac{1}{- 1 / 5}) {(e^{- t / 5})}_{0}^{10} = 125 (1 - e^{- 2})$
$= 125 (1 - \frac{1}{2 {.7}^{2}}) = 108 m$

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