A 5 kg projectile is fired at an angle of 25o above the horizontal. Its initial velocity is 200 m/s and just before it hits the ground its velocity is 150 m/s. What is the change in the mechanical energy of the projectile?

The change in the mechanical energy of the projectile is 43,750 J

Explanation:

Given;

mass of the projectile, m = 5 kg

initial velocity of the projectile, u = 200 m/s

final velocity of the projectile, v = 150 m/s

The change in mechanical energy is calculated from the principle of conservation of energy;

ΔP.E = ΔK.E

The change in potential energy is zero (0)

0 = ΔK.E

ΔK.E = K.E₁ - K.E₂

ΔK.E = ¹/₂mu² - ¹/₂mv²

ΔK.E = ¹/₂m(u² - v²)

ΔK.E = ¹/₂ x 5(200² - 150²)

ΔK.E =  43,750 J

Therefore, the change in the mechanical energy of the projectile is 43,750 J

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student