AP EAMCET202017 Sep 2020Morning ShiftPhysicsMotion in Two DimensionsActual
A particle is projected with a velocity (v ) such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is ((g= ) acceleration due to gravity)
Options
- A( 4 v^2 5 g )
- B( 4 g 5 v^2 )
- C( v^2 g )
- D( 4 v^2 5 g )
Correct answer
A. ( 4 v^2 5 g )
Step-by-step solution
Velocity of particle (=v ) If ( ) is angle of projection such a way, ( array ll & R=2 H & v^2 2 g = 2 v^2 ^2 2 g & 2 = ^2 & 2= 2= =2 array ) ( aligned & = 2 5 = 1 5 Range & = v^2 2 g = v^2 2 g & = v^2 2 2 5 1 5 g = 4 v^2 5 g aligned )