A particle is projected at an angle A and after t seconds, it appears to have an angle of B with the horizontal. The initial velocity is:- A)gt/(2 sin(A-B)) B)gt cos B/(sin (A-B)) C)sin(A-B)/2gt D)2 sin(A-B)/(gt cos B)?

This needs to be treated as motion of a Projectile along an inclined plane.

As shown in the figure above let O be the point of projection of the particle at an angle #A#. Let the particle strike the inclined plane having angle #B# with the horizontal, at A.
The initial velocity #vec{u}# can be resolved into two components.

1. along the plane #=u cos (A - B) #
2. perpendicular to the plane #=u sin (A - B) #

Similarly the acceleration due to gravity #g# can also be resolved into two components

1. parallel to the plane #=g sin B#
2. perpendicular to the plane #=g cos B#

Now time of flight #t# along the inclined plane can be found using the kinematic expression

#h=ut+1/2at^2#

When the particle goes from A to B, its displacement along the plane is zero. Considering components perpendicular to the plane, we get

#0 = u sin (A-B) t - 1/2(g cosB) t^2#
#=> u = (1/2(g cosB)t)/(sin(A-B))#
#=>u=(g t cos B)/(2sin (A-B))#

The acceleration vector is:

#bb a = ((0),(- g))#

#bb v = bb u + bba t = u((cos A),(sin A)) + ((0),(- g))t#

At time #t#, the angle #B# between the horizontal and the velocity vector follows from this:

#bb v * bb hat i = abs(bbv) cos Bimplies cos B = (bb v * bb hat i)/abs(bbv)#

And:

• #bb v * bb hat i = ((u cos A),(u sin A - g t)) *((1),(0)) = u cos A#

• #abs (bb v) = sqrt( ( u cos A)^2 + (u sin A - g t)^2 )#

# = sqrt( u^2 - 2 u g t sin A + (g t)^2 )#

#implies cos B = (u cos A)/(sqrt( u^2 - 2 u g t sin A + (g t)^2 ))#

#cos^2 B ( u^2 - 2 u g t sin A + (g t)^2 ) = u^2 cos^2 A#

# (cos^2 B - cos^2 A) color(red)(u^2) - 2cos^2 B sin A\ g t \ color(red)(u) + cos^2 B(g t)^2 = 0#

This is a quadratic. The discriminant - ie the #b^2 - 4ac # bit - is:

#4cos^4 B sin^2 A\ g^2 t^2 - 4(cos^2 B - cos^2 A) cos^2 B(g t)^2 #

#= (g t)^2 (4cos^4 B sin^2 A - 4cos^4 B + 4 cos^2 A cos^2 B)#

#= (g t)^2 (4cos^4 B( sin^2 A - 1) + 4 cos^2 A cos^2 B)#

#= (g t)^2 cos^2 A cos^2 B (- 4cos^2 B + 4 )#

#= (g t)^2 cos^2 A cos^2 B ( 4sin^2 B )#

#= 4 (g t)^2 cos^2 A cos^2 B sin^2 B #

The Quadratic Formula is therefore:

#u = ( 2cos^2 B sin A\ g t pm 2 g t cos A cos B sin B )/(2 (cos^2 B - cos^2 A))#

# =g t cos B ( cos B sin A\ pm cos A sin B )/( (cos 2 B + 1)/2 - (cos 2 A + 1)/2 )#

# =2 g t cos B ( sin (A pm B) )/( cos 2 B - cos 2 A )#

• #2 sin(A - B) sin(A + B) = cos 2 B - cos 2 A #

# =2 g t cos B ( sin (A pm B) )/( 2 sin(A - B) sin(A + B) )#

Either:

# =2 g t cos B ( sin (A + B) )/( 2 sin(A - B) sin(A + B) )#

# =(2 g t cos B )/( 2 sin(A - B) ) =( g t cos B )/( sin(A - B) )#

Or:

# =2 g t cos B ( sin (A - B) )/( 2 sin(A - B) sin(A + B) )#

# =( g t cos B )/( sin(A + B) )#

This second solution may reflect the projectile on descent. The first solution is squarely in the answer key.