The potential energy of a bullet weighing 0.05 kg, fired from the barrel vertically upwards

| June 15, 2021 | education

The potential energy of a bullet weighing 0.05 kg, fired from the barrel vertically upwards, 4 seconds after the start of movement is 40 J. What is the initial velocity of the bullet?

m = 0.05 kg.

g = 9.8 m / s2.

Ep = 40 J.

t = 4 s.

V0 -?

The potential energy of the body En is determined by the formula: En = m * g * h, where m is the mass of the body, g is the acceleration of gravity, h is the height of the body.

h = En / m * g.

h = 40 J / 0.05 kg * 9.8 m / s2 = 81.6 m.

Only gravity acts on the bullet, so it moves with the acceleration of gravity g.

Let’s find the initial bullet velocity V0 from the formula: h = V0 * t – g * t ^ 2/2.

V0 * t = h + g * t ^ 2/2.

V0 = h / t + g * t / 2.

V0 = 81.6 m / 4 s + 9.8 m / s2 * 4 s / 2 = 39.6 m / s.

Answer: the muzzle velocity of the bullet is V0 = 39.6 m / s.



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