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)?

2 Answers
May 26, 2018

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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))

May 27, 2018

( g t cos B )/( sin(A - B) )

Explanation:

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.