A ship maneuvers to within 2.50 x 10 3 m of anisland\'s 1.80 x 10 3 m high mount

Question

A ship maneuvers to within 2.50 x 103 m of anisland's 1.80 x 103 m high mountain peak and fires aprojectile at an enemy ship 6.10 x 102 m on teh otherside of the peak. If the ship shoots the projectile with an initialvelocity of 2.50 x 102 m/s at an angle of75.0o, how close to the enemy ship does the projectileland? How close (vertically) does the projectile come to thepeak? My crude drawing of the diagram they give you... A ship maneuvers to within 2.50 x 103 m of anisland's 1.80 x 103 m high mountain peak and fires aprojectile at an enemy ship 6.10 x 102 m on teh otherside of the peak. If the ship shoots the projectile with an initialvelocity of 2.50 x 102 m/s at an angle of75.0 degree, how close to the enemy ship does the projectileland? How close (vertically) does the projectile come to thepeak? My crude drawing of the diagram they give you...

Explanation / Answer

Now horizontal distance traveled by the projectile is R =vo2sin2 / g                = (2.50 x 102 m/s)2(sin150o) /(9.8m/s2)                = 3.188 x 103 m
Therefore the projectile lands at a distance of 3.188x103 m- 3.11 x 103 m= 78m closer to the enemy ship.
Now at a horizontal distance of 2.50 x 103 m,the vertical height of the projectile is given by
y = (tan )x - (1/2){g/[(vo)2(sin )2]}x2   = (tan75)(2.50 x 103 m)- (1/2){(9.8m/s2)/[(2.50 x 102 m/s)2(sin75o)2]}(2.50x 103 m)2   = 4.08 x 103 m
Then the projrctile is 4.08 x 103 m- 1.8 x 103 m= 2.28 x 103 mcloser to the peak of the mountain.

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