Assuming there is no atmosphere and the projectile is not aimed below the horizon (i.e. at the ground), the escape velocity IS independent of launch angle. Consider the situation from an energy point of view. If there is enough in initial kinetic energy so that Ve^2/2 = MG/R, where M is the Earth's mass and R is its radius, then when r = infinity, the total energy
(m/2)v^2 - mMG/r
is zero, and the projectile can escape. The horizontal component of velocity at launch IS used to overcome gravitational attraction at large distances.
How is escape velocity independent of angle of projection??
I mean i expect it to be DEPENDENT on the angle of projection cuz only vertical component of initial velocity helps a body to overcome gravitational force. So only the vertical component of velocity can behave as escape velocity. And angle does matter there.. PLzzzz help me out..
2 answers
break components of velocity and apply conservation of energy . You will observe that sin( theta ) and cos(theta) terms of velocity can be easily eliminated from sin square + cos square = 1. Hence we see that expression of escape velocity of object launched at angle theta from vertical will be same as that for vertical projection of body.