# Projectile motion of ship

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

Last edited:

lucianman24 said:

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?

THANK YOU

Hey, lucianman, welcome to the site!

What's your work so far for this problem? What values do you know? What equations might be useful?

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lucianman24 said:
P.S. Typing 'projectile motion' into the search box should be very useful too, since it is a highly frequent topic. 