A shot is fired at 30^o with the vertical from a point on the ground with kinetic energy K. If air resistance is ignored, the kinetic energy at the top of the trajectory is:

Question

Question

This question was previously asked in

ALP CBT 2 Physics and Maths Previous Paper 3 (Held On: 21 Jan 2019 Shift 3)

Answer 1

CONCEPT:

Projectile motion: Projectile motion is the motion of an object projected into the air, under only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.
- Initial Velocity: The initial velocity can be given as x components and y components.

ux = u cosθ

uy = u sinθ

Where u stands for initial velocity magnitude and θ refers to projectile angle.

Where h is the maximum height.

Where R is the total distance covered by the projectile.

Kinetic energy (KE) = (1/2)m V²

Where m is mass and V is the velocity

CALCULATION:

Given that:

(Angle with vertical = 30°)

θ = 60° (with horizontal 90o - 30o = 60o )

ux = u cosθ = u cos 60° = (1/2) u

uy = u sinθ = u sin 60° =√3/2 / 2

Kinetic energy on the ground (KE) = (1/2)m u2 = K

At the top, velocity = V = u / 2

Kinetic energy at top (KE') = (1/2)m (u/2)2 = mu²/8 = K/4

So option 1 is correct.