Question

The vody is projected at an angle of 30° with kinetic energy E . The kinetic energy at top most point is ?

Answer 1

Answer. :. Given that,

Given that,Angle θ=30

Given that,Angle θ=30 0

Given that,Angle θ=30 0Now, v is the velocity of projection

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy is

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE=

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 2

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30 0

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30 0 u=v×

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30 0 u=v× 2

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30 0 u=v× 23

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30 0 u=v× 23Now, kinetic energy at top

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30 0 u=v× 23Now, kinetic energy at top K.E=

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30 0 u=v× 23Now, kinetic energy at top K.E= 2

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30 0 u=v× 23Now, kinetic energy at top K.E= 21

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30 0 u=v× 23Now, kinetic energy at top K.E= 21

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30 0 u=v× 23Now, kinetic energy at top K.E= 21 m[

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30 0 u=v× 23Now, kinetic energy at top K.E= 21 m[ 2

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30 0 u=v× 23Now, kinetic energy at top K.E= 21 m[ 23

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30 0 u=v× 23Now, kinetic energy at top K.E= 21 m[ 23

Given that,Angle θ=30 0Now, v is the velocity of projectionThe energy isE= 21 mv 2At top the velocity u=vcos30 0 u=v× 23Now, kinetic energy at top K.E= 21 m[ 23 mv

v 2

v 2 ]

v 2 ] 2

v 2 ] 2

K.E=

K.E= 4

K.E= 43

K.E= 43

K.E= 43 (

K.E= 43 ( 2

K.E= 43 ( 21

K.E= 43 ( 21

K.E= 43 ( 21 mv

K.E= 43 ( 21 mv 2

K.E= 43 ( 21 mv 2 )

K.E= 43 ( 21 mv 2 ) K.E=

K.E= 43 ( 21 mv 2 ) K.E= 4

K.E= 43 ( 21 mv 2 ) K.E= 43E

K.E= 43 ( 21 mv 2 ) K.E= 43E

K.E= 43 ( 21 mv 2 ) K.E= 43E

K.E= 43 ( 21 mv 2 ) K.E= 43E Hence, the kinetic energy is

K.E= 43 ( 21 mv 2 ) K.E= 43E Hence, the kinetic energy is 4

K.E= 43 ( 21 mv 2 ) K.E= 43E Hence, the kinetic energy is 43E

K.E= 43 ( 21 mv 2 ) K.E= 43E Hence, the kinetic energy is 43E

K.E= 43 ( 21 mv 2 ) K.E= 43E Hence, the kinetic energy is 43E at top