Question

A projectile is fired at an angle theta with the horizontal.find the condition under which it lands perpendicular on an inclined plane of inclination Alpha

Answer 1

let b be the angle of projection with the inclined plane whose inclination is a

then the condition for it to land perpendicular the plane is 2tan(b-a)=cota

if the projectile land on the surface horizontally(parallel to ground ) then tanb = cota+2tana

it can be easily derived from the formulas of motion on inthe inclined plane such as

x=ucos(b-a)t -(1/2)gsin(a)t^2

y= usin(b-a)t-(1/2)gcos(a)t^2

by appling the condition

I HOPE THIS ANSWER IS HELPFULL FOR YOU...

Answer 2

Let angle of inclination be a

And angle of projection with the horizontal be b.

We know,

y=u.sin (b-a)t-1/2gcos a t^2

y=0

0=u.sin (b-a)t-1/2gcos a t^2

u.sin (b-a)t = 1/2gcos a t^2

u.sin (b-a) = 1/2gcos a t

t=2usin(a-b)/gcosa.

tan 90°= Vy/Vx.

1/0=Vy/Vx

Vx=0

u.cos (b-a)t-g sina t=0

u.cos (b-a)t = g sina t

Putting t=2usin(a-b)/gcosa

u.cos (b-a)t = g sina2usin(a-b)/gcosa

Cot(a-b)= 2 tan a.

Hope this will help you..☺️