Question

A particle is projected horizontally with speed u from point A, which is 10 m above the inclined plane perpendicularly at point B. $[g=10m/s^2]$

Find horizontal speed with which the particle was projected

Answer 1

Taking +ve x-axis parallel to OB and y-axis perpendicular to it.
$u_x=ucos{45}^{\circ }$
$u_y=usin{45}^{\circ }$
$a_x=-gsin{45}^{\circ }$
$a_y=gcos{45}^{\circ }$
To determine ''t''
$v_x=u_x+a_{x}t$
$v_x=0$
$t=-\left(\frac{u_x}{a_x}\right)$
Or, $t=\frac{ucos{45}^{\circ }}{gsin{45}^{\circ }}$
$t=\frac{u}{g}$
Length of the projection is, $l=10sin{45}^{\circ }$
$y=u_y+a_y\frac{t^2}{2}$
$10sin{45}^{\circ }c=usin{45}^{\circ }+\frac{1}{2}gcos{45}^{\circ }{\left(\frac{u}{g}\right)}^2$
$10=u+\frac{u^2}{2g}$
Or $u^2+2ug=20g$
$u^2+20u-200=0$
On solving the quadratic equation, you can calculate the value of ''u''