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

• A. $2s$
• B. $2\sqrt{10}$ s
• C. $10\sqrt{2}$ s
• D. $\frac{1}{2}$ s

Answer 1

The correct option is A $2s$

Let t be the time of passing between the walls

$S=vt$

$20=20cos\theta t$ -------------(i)

vertically

$v^2=u^2+2aS$

$v^2=u^{2}si{n}^2\theta -2g\times 10$

$v^2=u^{2}si{n}^2\theta -20g$

$v=u+at$

$0=u-gt$

$t=\frac{v}{g}$------------------(ii)

equating eq i and ii

$\frac{1}{cos\theta }=2\frac{v}{g}$

squaring both sides and substituting v

$\frac{g^2}{co{s}^2\theta }=4u^{2}si{n}^2\theta -80g$

let $si{n}^2\theta$ be x

$g^2=16g^{2}x(1-x)-8g^2(1-x)$

$9=16x-16x^2+8x$

$16x^2-24x+9=0$

$(4x-3{)}^2=0$

$x=\frac{3}{4}$

$si{n}^2\theta =\frac{3}{4}$

$\theta =\frac{\pi }{3}$

therefore time of passing between the walls

$t=\frac{1}{cos\theta }=2s$