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Projectile motion of ship

  1. Oct 22, 2006 #1
    Please help me with the following question:

    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: Oct 22, 2006
  2. jcsd
  3. Oct 22, 2006 #2
    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?
  4. Oct 22, 2006 #3


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    Show us some work. You should start off by writing down all the equations related to projectile motion (i.e. velocity, displacement, etc.).

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