Question

a particle is projected horizontally with speed u from point A,which is 10m above the ground if the particle hits the inclined plane perpendicularly at point B
Find the horizontal speed with which the particle was projected

Answer 1

Final Answer : $10 \sqrt{\frac{2}{3}} m/s$

Steps:
1) Look at the figure to watch geometrical manipulations.
Now,
When ball strikes the inclined plane, it makes an angle of 45° with the horizontal.
Let it strikes the inclined plane at time 't' s.
Then, horizontal velocity remains constant.
$v_{x@t} = u$
Then,
$tan(45\degree) = \frac{v_x}{v_y} \\ \\ => 1 = u/v_y \\ \\ => v_y = u$

2) Then,
Convention : Downward and right directions are positive.

$v_y = u_y + at \\ \\ => u = 0 + gt \\ \\ => t = \frac{u}{g}$

3) Then,
Using
$tan(45\degree) = \frac{BD}{OD} \\ \\ => BD = OD \\ \\ => u_x t = 10-S_y \\ \\ => ut = 10- \frac{1}{2}gt^2 \\ \\ => \frac{u^2}{g} = 10- \frac{1}{2}g(u/g)^2 \\ \\ => \frac{u^2}{10} + \frac{u^2}{20} = 10 \\ \\ => u =10 \sqrt{\frac{2}{3}}$

Therefore, Horizontal Speed of particle is
$u =10 \sqrt{\frac{2}{3}} m/s$