Question

A particle is projected horizontally with a speed u from the top of a plane inclined at an angle ${\theta }^0$ with the horizontal. How far from the point of projection will the particle strike the plane?

Answer 1

Take X,Y axes as shown in fugure. Suppose that the particle strikes the plane at a point P with coordinates (x,y). Consider the motion between A and P.

Now,

Motion in x direction $=u$

Acceleration $a=0$

And $x=ut$

Now

Motion in y direction

Initial velocity $u=0$

Acceleration $a=g$

So,

$y=ut+\frac{1}{2}g{t}^2=\frac{1}{2}g{t}^2$

$y=\frac{!}{2}g{t}^2$

Now eliminate t from x and y respectively, we get

$y=x\tan \theta ..........\left(1\right)$

Thus,

$x\tan \theta =\frac{g{t}^2}{2u^2}$

On solving for x get two values i.e.

$x=0$ and $x=\frac{2u^2\tan \theta }{g}$

Therefore the point is corresponding to $x=\frac{2u^2\tan \theta }{g}$

So put this value of x in equation (1) we get,

$y=x\tan \theta =\frac{2u^2{\tan }^2\theta }{g}$

Then,

The distance is $OP=\sqrt{x^2+y^2}$

$=\frac{2u^2\tan \theta \sqrt{1+\tan \theta }}{g}$

$=\frac{2u^2\tan \theta \sec \theta }{g}$