Question

projectile fired at an angle of 30° with horizontal.Its kinetic energy at initial is E,find the kinetic energy at max height​

Answer 1

Answer:

3E/4

Explanation:

The kinetic energy at highest point is given as :

K.E. = (cos α)2 E

α = 30°

K.E. = (cos 30)2 E

K.E. = (√3/4)2 E

= (3/4) E

Answer 2

Answer : (3/4) E

Initially, when projectile is fired with an angle of 30° with horizontal it has both vertical and horizontal velocities which contribute to its total kinetic energy E.

However, at maximum height it's vertical velocity is zero and only horizontal velocity contributes to its total kinetic energy E'.

Let projectile be fired with velocity u making an angle of 30° with horizontal

Horizontal component of velocity

= u cos x

Vertical component of velocity

= u sin x

where x = 30°

Initial Kinetic energy

= 1/2 m (u cos x)^2 + 1/2 m (u sin x)^2

= 1/2 m (u)^2 = E

because, (sin x)^2 + (cos x)^2 = 1

Final Kinetic Energy

= 1/2 m (u cos x)^2 =E'

E/E' = {1/2 m (u)^2} ÷ {1/2 m (u cos x)^2}

E/E' = (u)^2÷(u cos x)^2

E/E' = u^2 ÷ u^2 (cos x)^2

E/E' = 1÷(cos x) ^2

= 1÷(cos30°)^2 [x=30°]

= 1÷(√3/2)^2

= 1÷(3/4)

= 4/3

E/E' = 4/3

E = E' × 4/3

E' = E × 3/4

THEREFORE,

E' = (3/4)E