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
ugtopper.blogspot.in edited
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
The initial velocity
- along the plane
=u cos (A - B) - perpendicular to the plane
=u sin (A - B)
Similarly the acceleration due to gravity
- parallel to the plane
=g sin B - perpendicular to the plane
=g cos B
Now time of flight
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))
Explanation:
The acceleration vector is:
At time
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 )
This is a quadratic. The discriminant - ie the
The Quadratic Formula is therefore:
2 sin(A - B) sin(A + B) = cos 2 B - cos 2 A
Either:
Or:
This second solution may reflect the projectile on descent. The first solution is squarely in the answer key.