Question

Find the relation between alpha and beta such that object strikes the inclined plane normally​

Answer 1

see the attachment mate ✌️☺️

Answer 2

The trajectory equation of a projectile is -

y=xtan(α)−gx22u2Cos2(α)

Let us find the point where the inclined plane intersect the trajectory.

Equation of inclined plane : y=tan(30)x

y=13–√x ——— eqn 1

Solving this with the trajectory equation to find the point where particle lands on the plane. Put eqn 1 in the trajectory eqn -

13–√x=tan(α)x−gx22u2Cos2(α)

(13–√−tan(α))x+(g2u2Cos2(α))x2=0

x((13–√−tan(α))+(g2u2Cos2(α))x)=0

(13–√−tan(α))+(g2u2Cos2(α))x=0

x=(3–√tan(α)−1)(2u2Cos2(α))3–√g

This is the point where the particle lands on the plane, perpendicularly. Hence the slope of tangent drawn to trajectory at this point must be perpendicular to plane, i.e. must make angle -60°, with the X-axis. Thus the slope of tangent to trajectory must be tan(-60°). We can find the slope of tangent by differentiating the trajectory equation -

dydx|xintersection=tan(−60°)

ddx(tan(α)x−gx22u2Cos2(α))=−3–√

tan(α)−gxu2Cos2(α)=−3–√

Put the value of xintersection calculated earlier here -

tan(α)−gu2Cos2(α)(3–√tan(α)−1)(2u2Cos2(α))3–√g=−3–√

tan(α)−2(3–√tan(α)−1)3–√=−3–√

3–√tan(α)−23–√tan(α)+2=−3  

3–√tan(α)=5

tan(α)=53–√

α=tan−1(53–√)