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<p>A ship maneuvers to within 2.46 * 10³m of an island's 1.78 * 10³m high mountain peak and fires a projectile at an enemy ship 6.05 * 10²m on the other side of the peak as illustrated.&#160; The ships shoots the projectile with an initial velocity of 2.52 * 10²m/s at an angle of 70.7 degrees.&#160; The acceleration of gravity is 9.81 m/s^{2.&#160;&#160;}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

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