Aship maneuvers to within 2.50 x 103 m of an island's 1.78 x 103 m high mountain peak and fires a projectile at an enemy ship 6.09 x 102 m on the other side of the peak. the ship shoots the projectile with an initial velocity of 2.54 x 102 m/s at an angle of 74.9 degress. the acceleration of gravity is 9.81 m/s2.

a) how close to the enemy ship does the projectile land? answer in units of m

b) how close (vertically) does the projectile come to the peak? answer in unity of m

a)

m

b)

m

Explanation:

a)

= initial velocity of launch = 254 ms⁻¹

= Angle of launch = 74.9 deg

Consider the motion of the projectile along the vertical direction

= initial velocity of projectile along vertical direction =  = = 245.23 ms⁻¹

= acceleration = - 9.8 ms⁻²

=vertical displacement = 0 m

= time of travel = ?

Using the kinematics equation

inserting the values

s

Consider the motion of the projectile along the horizontal direction

= initial velocity of projectile along horizontal direction =  = = 66.17 ms⁻¹

= acceleration = 0 ms⁻²

= horizontal displacement

= time of travel = 50.05 s

Using the kinematics equation

m

= distance of the enemy ship from the point of launch = 2500 + 609 = 3109 m

= closest distance from the ship

m

b)

Consider the motion of the projectile along the horizontal direction from the point of launch to till it reach the peak of mountain

= initial velocity of projectile along horizontal direction =  = = 66.17 ms⁻¹

= acceleration = 0 ms⁻²

= horizontal displacement = 2500 m

= time of travel = ?

Using the kinematics equation

s

Consider the motion of the projectile along the vertical direction from the point of launch to till it reach the peak of mountain

= initial velocity of projectile along vertical direction =  = = 245.23 ms⁻¹

= acceleration = - 9.8 ms⁻²

= vertical displacement = ?

= time of travel = 37.78 s

Using the kinematics equation

m

= height of the mountain = 1780 m

= the closest distance of projectile from peak

m