A particle is projected with velocity 20m/s, so that it just clears two walls of equal height 10m, which are - Physics - Physical World

Question

A particle is projected with velocity?20m/s, so that it just clears
two walls of equal height?10m, which are at a distance?20m?from
each other The time of passing between the walls is

Deepak Kumar · Accepted Answer

Dear Student ,
Let x be the distance of the first wall from the point where the particle was projected.

$x + 2 h = v \cos θ t_{2} x = v \cos θ t_{1} \Rightarrow t_{2} - t_{1} = \frac{2 h}{v \cos θ} h = v \sin θ t - \frac{1}{2} g t^{2} = v \sin θ t - 5 t^{2} \Rightarrow 5 t^{2} - v \sin θ t + h = 0 t_{1} + t_{2} = \frac{v \sin θ}{5} a n d t_{1} t_{2} = \frac{h}{5} \Rightarrow {(t_{1} - t_{2})^{2} = (\frac{v \sin θ}{5})}^{2} - \frac{4 h}{5} \Rightarrow \frac{4 h^{2}}{v^{2} \cos^{2} θ} = \frac{v^{2} \sin^{2} θ}{25} - \frac{4 h}{5}$

$\frac{4 h^{2}}{v^{2} \cos^{2} θ} = \frac{v^{2} \sin^{2} θ}{25} - \frac{4 h}{5} v^{2} = 4 g h \Rightarrow \frac{h}{10 \cos^{2} θ} = \frac{8 h \sin^{2} θ}{5} - \frac{4 h}{5} \Rightarrow \frac{1}{2 \cos^{2} θ} = 8 \sin^{2} θ - 4 \Rightarrow \cos θ = \frac{1}{2} t_{2} - t_{1} = \frac{2 h}{v \cos θ} = \frac{4 h}{2 \sqrt{10 h}} = \sqrt{\frac{2 h}{5}}$
given that h = 10 m
The time of passing between the walls is = 2sec
So option (1) is correct
Regards

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A particle is projected with velocity?20m/s, so that it just clears two walls of equal height?10m, which are at a distance?20m?from each other. The time of passing between the walls is