The kinetic energy of a body weighing 1 kg, thrown horizontally at a speed of 30 m / s 4 s after the fall is equal.

| September 23, 2021 | education

m = 1 kg.

V0 = 30 m / s.

t = 4 s.

g = 10 m / s2.

Ek -?

The kinetic energy of the body Ek is called the product of the body’s mass m by the square of the speed of its movement V2: Ek = m V2 / 2.

Horizontally, the body will move uniformly at a speed V0. Will fall free vertically. This means that it will move uniformly with the acceleration of gravity g.

The vertical component of the velocity V1 will have the value V1 = g * t = 10 m / s2 * 4 s = 40 m / s.

The body has two mutually perpendicular velocities, the horizontal component V0, and the vertical component V1. By the Pythagorean theorem, the resulting velocity of the body V is expressed by the formula: V = √ (V02 + V12).

V = √ ((30 m / s) 2 + (40 m / s) 2) = 50 m / s.

Ek = 1 kg * (50 m / s) 2/2 = 1250 J.

Answer: the body will have kinetic energy Ek = 1250 J.



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