Answer to Question #86791 in Mechanics | Relativity for Abel

Question

The attacking ship has manoeuvred to 2500 m distance away from the mountain peak (1800 m high) and can shoot projectiles with an initial speed of 250 m/s. The shoreline on the right side of the mountain is horizontally 300 m away from its peak. Then the attacking ship has manoeuvred to take a new position to be at 2250 m away from the mountain peak. What are the distances right side of the mountain at which the defending ship can be safe from the bombardment of the attacking ship?

Answer 1

By understanding the problem, the attacking ship is ultimately at the 2250 m distance from the mountain peak (horizontally).

We will denote the given values as:

$x_a=2250~\text{m};~h=1800~\text{m};~x_{shore}=300~\text{m};~v=250\frac{\text{m}}{\text{s}}. \\$

The trajectory of a projectile motion:

$y(x) = x\tan\alpha - \frac{gx^2}{2v^2\cos^2\alpha}. \\$

By making sure that by the moment when a projectile reaches the mountain, it can pass over its peak, we can find the acceptable range of angles:

$y(x_a) = x_a\tan\alpha - \frac{gx_a^2}{2v^2\cos^2\alpha} > h; \\$

We shall enter numerical values and resort to some calculation aid here:

In: Solve[2250*Tan[a] - (9.81*(2250)^2)/(2*((250)^2)*(Cos[a])^2) == 1800, a]
Out: {{a -> -2.2438}, {a -> -1.79385}, {a -> 0.897791}, {a -> 1.34775}}

By discarding the negative results and converting the results to degrees (for better illustration):

In: 0.8977911361271153*180/Pi
Out: 51.4396

In: 1.347746132891334*180/Pi
Out: 77.2202

we can see that projectile will pass over the peak if shot between the angles of 51.44 and 77.22 degrees.

Simple check of the maximum range $L = \frac{v^2}{g}\sin{(2\alpha)}$ will prove that the angle 51.44 degrees will give farther range:

In: (250^2)*Sin[2*0.8977911361271153]/9.81
Out: 6210.77

In: (250^2)*Sin[2*1.347746132891334]/9.81
Out: 2748.79

Therefore let's label $\alpha_m \approx 51.44\degree. \\$

Maximum range, as already calculated, $L_{max} = \frac{v^2}{g}\sin{(2\alpha_m)} \approx 6211~\text{m}.$

So the safe distance for the defending ship, if calculating from the peak base, horizontally is

$x_{safe} = L_{max} - x_a \approx (6211-2250)~\text{m} \approx 3961~\text{m}. \\$

One can also obviously see that the projectile in this case will pass the horizontal distance of the shore (300 m).