Question

A bullet of mass 10 gram is fired horizontally with a velocity 1000m per second from rifle situated at a height 50 m above the ground if the bullet reaches the ground with the velocity 500 m per second the work done against air resistance in the trajectory of the bullet is

Answer 1

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Answer 2

The work done against air resistance in the trajectory of the bullet is 3754.9J

When the bullet is fired with initial velocity 'u', its Kinetic energy would be:

(1/2)mu², where 'm' is the mass of the bullet.

Also potential energy at the height 50m above ground would be mgh.

So, total initial energy = (1/2)mu² + mgh

When the bullet reaches the ground, its potential energy is 0 (as height from ground is 0)

So final energy = Kinetic energy = (1/2)mv², where 'v' is the final velocity.

Work done against air resistance = Total Initial Energy - Final energy

= (1/2)mu² + mgh - (1/2)mv²

= [(1/2)×(10×10⁻³)×(10⁶)] + [(10×10⁻³)×(9.8)×(50)] - [(1/2)×(10×10⁻³)×(0.25×10⁶)]

= 3754.9J                                          [1 g = 10⁻³kg and g = 9.8m/s²]