Question

For a particle projected with a speed 20 m/s from ground, the
kinetic energy at top most point on the trajectory is of the
initial kinetic energy. The angle of projection from the
horizontal is
30
45
60°
37​

Answer 1

Explanation:

For a particle projected with a speed 20 m/s from ground, the

kinetic energy at top most point on the trajectory is of the

initial kinetic energy. The angle of projection from the

horizontal is

30

Answer 2

Concept:

• Projectile motion
• Applying kinematics equations

Given:

• initial speed u = 20m/s
• Kinetic energy at the topmost point = 1/4th of initial kinetic energy

Find:

• The angle of projection from the horizontal

Solution:

At the topmost point, the vertical component of velocity is zero.

There is only the horizontal component of the velocity.

The horizontal component of velocity = u cos θ

Kinetic energy at topmost point = 1/2m(u cos θ)^2

Initial kinetic energy = 1/2mu^2

Kinetic energy at the topmost point = 1/4th of initial kinetic energy

1/2m(u cos θ)^2 = (1/4) 1/2mu^2

(u cos θ)^2= u^2/4

(cos θ)^2 = 1/4

cos θ = 1/2

θ = 60°

The angle of projection from the horizontal is 60°.

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