Projectile Motion

Question

During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 1.75 km/s at an initial inclination of 27◦ to the horizontal. The acceleration of gravity is 9.8 m/s 2 . How far away did the shell hit? Answer in units of km. How long was it in the air? Answer in units of s.

Answer 1

Let's start with our kinematic equations:

x = v₀cos(θ)*t

y = v₀sin(θ)*t - (1/2)gt²

Since we are solving for a parabola with roots where y = 0, we can find the time(s) t where y equals 0. We can then use t to satisfy the equation for the range, x.

1) y = 0

2) v₀sin(θ)*t = (1/2)gt² - > v₀sin(θ) = (1/2)gt -> t = 2v₀sin(θ)/g

3) x = 2v₀²sin(θ)cos(θ)/g = v₀²sin(2θ)/g

Now, we can solve the range equation using the given values. It will be handy to convert the acceleration due to gravity to km/s², since the desired answer is in units of km.

g = 9.8*10^-3 km/s²

4) x = (1.75 km/s)² *sin(54°) / (9.8*10^-3 km/s²) = 252.8 km ≈ 253 km

Using our approximate solution, we can calculate air time. The easiest way to do this is to just use the equation for range and rearrange for time.

5) t = x/(v₀cos(θ)) = 253 km / ((1.75 km/s)*cos(27°)) = 162.2 s ≈ 162 s

See that the projectile travels about 253 kilometers in about 2 minutes and 42 seconds