An enemy ship is on the east side of a mountain island, as shown in the figure.
ID: 1771430 • Letter: A
Question
An enemy ship is on the east side of a mountain island, as shown in the figure. The enemy ship has maneuvered to within d1 = 2020 of the h = 1640 m high mountain peak and can shoot projectiles with an initial speed of vi = 242 m/s. If the western shoreline is horizontally d2 = 299 m from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship?
An enemy ship is on the east side of a mountain island, as shown in the figure. The enemy ship has maneuvered to within d1 - 2020 m of the h 1640 m high mountain peak and can shoot projectiles with an initial speed of v, 242 m/s. If the western shoreline is horizontally d2 = 299 m from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship? di less thar m, or more thanExplanation / Answer
This can be seen as a problem of projectile motion. The initial launch projectile velocity is given as vi=240 m/s.
The maximum height of the projectile that can be launched from the enemy ship to cross the peak is given as h=1640 m.
The maximum height of the projectile is given as :
vi^2 sin^2(H)/(2g)=1640 m.
We know vi, g=9.81 m/s^2. Hence plugging these values in the above equation and solving for H we get H=32 degrees.
The range of the projectile is given as :
vi^2 sin(2H)/(g). We now know all these values. Hence 240*240*sin(2*32)/9.81 = 5268.684 m.
This is the distance from the enemy ship launching the missile from the eastern shore. The ship is (2020+299) m from the western shore. Hence the distance of between the western shore and the location of the projectile landing is 5268.684-(2020+299)=2949.684 m. Hence the ship on the western shore is safe within 2949.684 m from the western shore.
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