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A ship maneuvers to within 2.50 x 10 ^ 3 m of an island's 1.80 x 10 ^ 3 m high mountain peak and fires a projectile at an enemy ship 6.10 x 10 ^ 2 m on the other side of the peak. If the ship shoots the projectile with an initial velocity of 2.50 x 10 ^ 2 m/s at an angle of 75 degrees, how close to the enemy ship does the projectile land? How close (vertically) does the projectile come to the peak? My physics book says the answer will be 8m and 210m but I want to know how to do it.

Vertical Motion of a Projectile that falls from rest

Vy,f = -gt (where g = a = 9.81) (t = time)

Vy,f^2 = -2gy (y = delta y= displacement in the y direction)

delta y = -1/2g(t)^2

Horizontal Motion of a Projectile

Vx = Vx,i = constant ( i = intial velocity)

delta x = Vxt ( delta x = dispalcement in the x direction)

Projectiles Launched At An Angle

Vx = Vi (cos theta) = constant (theta = degrees of angle)

delta x = Vi (cos theta)t

Vy,f = Vi (sin theta) - gt

Vy,f^2 = Vi^2 (sin theta)^2 - 2g(delta y)

delta y = Vi (sin theta)t - 1/2g(t)^2

Hope you can understand if not reply saying what you dont

Thank You

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# Homework Help: Projectile motion of ship

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