Sship maneuvers to within 2.50x10^3 m of an islands 1.80x10^3 m high mountain peak and fires a projectile at an enemy ship 6.10x10^2 m on the other side of the peak. if the ship shoots projectile with an initial velocity of 2.50x10^2m/s at an angle of 75 degrees, how close vertically does the projectile come to the peak

The projectile traveled past the height of the enemy's ship.

The given parameters;

The maximum height traveled by the projectile is calculated as follows;

Total vertical height traveled by the projectile is 2,975.63 m.

The position of the enemy's ship on the other side of the peak = 610 m

Thus, we can conclude that the projectile traveled past the height of the enemy's ship.

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velocity of the projectile is given as

now the angle is given as 75 degree

so here we will have

now the time taken to reach the peak is given as

now the height moved by the projectile in same time is given as

now we have

so distance between the peak and projectile is given as

answer; i believe the correct answer is (40,000 j) good luck

the answer would be c