The citizens of Paris were terrified during World War I when they were suddenly bombarded with shells fired from a long-range gun known as Big Bertha. The barrel of the gun was 36.6 m long, and it had a muzzle speed of 2.20 km/s. When the gun’s angle of elevation was set to 55.0°, what would be the range? For the purposes of solving this problem, ignore air resistance. (The actual range at this elevation was 121 km; air resistance cannot be ignored for the high muzzle speed of the shells.)

464.1 km

Explanation:

Using R = u²sin2Ф/g where R = range of projectile, u = muzzle speed = 2.20 km/s = 2.20 × 10³ m/s, Ф = angle of elevation or projection angle = 55° and g = acceleration du° to gravity = 9.8 m/s²

Substituting the values of the variables into the equation, we have

R = u²sin2Ф/g

R = (2.20 × 10³ m/s)²sin2(55°)/9.8 m/s²

R = 4.84 × 10⁶ m²/s²sin110°/9.8 m/s²

R = 4.84 × 10⁶ m²/s² × 0.9397/9.8 m/s²

R = 4.548 × 10⁶ m²/s²/9.8 m/s²

R = 0.4641 × 10⁶ m

R = 464.1 × 10³ m

R = 464.1 km

answer: he got 23

explanation: