Question

A 5-kg projectile is fired over level ground with a velocity of 200 m/s at an angle of 25

above the horizontal. Just before it hits the ground its speed is 150 m/s. Over the entire trip the
change in the thermal energy of the projectile and air is:

Answer 1

Answer: 43.8 kJ

Step-by-step explanation:

Given;

mass of the object, m = 5kg

initial velocity of the projectile, v₁ = 200 m/s

final velocity of the projectile, v₂ = 150 m/s

To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.

Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²

KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)

KE = ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ

Therefore, the change in thermal energy of the projectile is 43.8 kJ