Question
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<p>A ship maneuvers to within 2.46 * 10<sup>3 </sup>m of an island's 1.78 * 10<sup>3 </sup>m high mountain peak and fires a projectile at an enemy ship 6.05 * 10<sup>2 </sup>m on the other side of the peak as illustrated.  The ships shoots the projectile with an initial velocity of 2.52 * 10<sup>2 </sup>m/s at an angle of 70.7 degrees.  The acceleration of gravity is 9.81 m/s<sup>2.   </sup>How close to the enemy ship does the projectile land and how close (vertically) does the projectile come to the peak?</p>
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Explanation / Answer
Given Distance between maneuvers ship and mountain is x1 = 2460 m Height of the mountain is yo = 1780 m Distance between enemy ship and mountain is x 2 = 605 m Initial velocity of projectile is u = 252 m/s Angle of projection = 70.7 o Acceleration due to gravity g = 9.81 m/s2 Horizontal distance travelled by the projectile is
R =
u 2 sin 2 /
g = ( 252 m/s ) 2 sin ( 2* 70.7 ) / 9.8 m/s 2 = 4042.739 m Close to the enemy ship does the projectile land is
x =
R - (
x1 +
x 2 ) = 4042.739 m - ( 2460 + 605 ) m = 977.739 m ______________________________________________________ Horizontal distance travelled by the projectile from maneuvers ship to mountain is
x 1 =
u cos *
t t =
x 1 /
u cos = 2460 m / ( 252 m/s )cos 70.7 = 29.535 s Vertical distance is
y =
u sin
t - ( 1/2)
g t 2 = (252 m/s ) (sin 70.7)( 29.535 s ) - ( 0.5 ) ( 9.81 m/s2 ) ( 29.535)2 . = 2745.82 m close (vertically) does the projectile come to the peak
y =
y -
y o = 2745.82 - 1780 m = 965.82 m ______________________________________________________ Horizontal distance travelled by the projectile from maneuvers ship to mountain is
x 1 =
u cos *
t t =
x 1 /
u cos = 2460 m / ( 252 m/s )cos 70.7 = 29.535 s Vertical distance is
y =
u sin
t - ( 1/2)
g t 2 = (252 m/s ) (sin 70.7)( 29.535 s ) - ( 0.5 ) ( 9.81 m/s2 ) ( 29.535)2 . = 2745.82 m close (vertically) does the projectile come to the peak
y =
y -
y o = 2745.82 - 1780 m = 965.82 m