Question

From the foot of an inclined plane of inclination $\alpha$, a projectile is shot at an angle $\beta$ with the inclined plane. Find the relation between $\alpha$ and $\beta$ if the projectile strikes the inclined plane perpendicularly to the plane.

• A. $2\tan \beta =\cot \alpha$
• B. $\tan \beta =\cot \alpha$
• C. $2\tan \beta =3\cot \alpha$
• D. $\tan \beta =2\cot \alpha$

Answer 1

The correct option is A $2\tan \beta =\cot \alpha$

In this question

first Let us assume that that that velocity of the projectile at the point of inclination be v

so we break the v as in the direction parallel to the inclined and perpendicular to the inclined plane that are $v\cos \beta$ and $v\sin \beta$

so we know that the projectile strikes the inclines plane perpendicular to the surface so the velocity of the projectile along the plane should be zero .

now acceleration the projectile in the direction perpendicular to the plane is $g\cos \alpha$

acceleration the projectile in the direction parallel to the plane is $g\sin \alpha$

as the velocity of the projectile along the plane should be zero when it strike the plane

we know the time period of a projectile or we can get it as the displacement of the projectile is zero in dirction perpendicular to the inclined plane

so $0=\left(v\sin \beta \right)t-\frac{1}{2}\left(g\cos \alpha \right)t^2$

so $T=\frac{2v\sin \beta }{g\cos \alpha }$

now we know that the velocity of the projectile along the plane should be zero

so we will use parallel to the plane

$v=u+a_{||}T$

so here $u=g\cos \beta$ and $a_{||}=-g\sin \alpha$ and T as calculated above and v = 0

so

so by solving this

so

option (a)