Question

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 1.10 km/s. when the guns 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.)
□ km

Explanation:

Step1: Convert speed to m/s

Muzzle speed $v_0 = 1.10\ \text{km/s} = 1100\ \text{m/s}$

Step2: List known values

Gravitational acceleration $g = 9.8\ \text{m/s}^2$, elevation angle $\theta = 55.0^\circ$

Step3: Apply projectile range formula

The range formula for projectile motion (flat ground, no air resistance) is $R = \frac{v_0^2 \sin(2\theta)}{g}$
Substitute values:

$$ R = \frac{(1100)^2 \times \sin(2\times55.0^\circ)}{9.8} $$

First calculate $2\theta = 110.0^\circ$, $\sin(110.0^\circ) \approx 0.9397$

$$ R = \frac{1210000 \times 0.9397}{9.8} \approx \frac{1137037}{9.8} \approx 116024.18\ \text{m} $$

Step4: Convert range to km

$$ R = \frac{116024.18}{1000} \approx 116\ \text{km} $$

Answer:

116 km