How to Derive the Inclination Angle of a Projected Particle?

[sara_87]

Homework Statement

A particle is projected at an angle alpha above the horizontal from a point on the edge of a table of height h standing on a horizontal floor. The particle reaches the floor at a point whose horizontal distance from the point of projection is 2h. Show that when it strikes the floor, the inclination theta below the horizontal of its direction of motion is given by

tan(theta) = tan(alpha) + 1

The Attempt at a Solution

okay i can manage to prove upto :

tan(theta) = tan(alpha) - gt/vi cos(alpha)


but i don't know what to do next... or rather what i do next doesn't make the fraction part equal 1

help would be much appreciated

 

[sara_87]

this might help...

is the method correct?

 

[Shooting Star]

(I was not quite able to read all your calculations, so I’ll give you a derivation my way.)

I’m writing ‘a’ for alpha and ‘b’ for theta, for convenience. The total time is t. Also, I’m writing u for the initial velo and v for the final velo.

u(cos a)t = 2h => t = 2h(ucos a) ..(1).
[Also, u cos a = v cos b, since they’re horz components.]

v^2 = u^2 + 2gh => h = (v^2 –u^2)/2g ..(2)

The eqn for y is: y = u(sin a)t – (1/2)gt^2. Putting (-h) in place of y,

-h = u(sin a)t – (1/2)gt^2
-h = (u*sin a)2h/ucos a –(1/2)g *4h^2/(u^2 cos^2 a) (from 1 putting value of t)
-1 = 2tan a – 2gh/(ucos a)^2
-1 = 2tan a – [2g/(ucos a)^2]( v^2 –u^2)/2g (from 2 putting value of h)
-1 = 2tan a – (v^2 –u^2)/(ucos a)^2
-1 = 2tan a – v^2/(ucos a)^2 – u^2)/(ucos a)^2
-1 = 2tan a – v^2/(vcos b)^2 + u^2)/(ucos a)^2 (since ucos a = vcos b)
-1 = 2tan a – sec^2 b + sec^2 a [put sec^2 = 1 + tan^2]
tan^2 b = (tan a + 1)^2, after some simplification. So,

tan(theta) = tan(alpha) + 1, taking the +ve value.