An enemy ship is on the east side of a mountainous island. The enemy ship can maneuver to within 2500 meters of the 1800 meter high mountain peak and can shoot projectiles with an initial speed of 250m/sec. If western shoreline is horizontally 300 meters from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship?
Draw the figure. Figure the max range is at a shooting angle of 45 deg.
Determine initial vertical vertical velocity (250)sin45
Deterine the time in air:
finalvertical position = initial vertical position + initialverticalvelocity*time + 1/2 (g)time^2
the initial and final vertical postion is zero, so
250sin45t - 4.9 t^2=0
solve for time in air.
Now, your horizontal equation.
xfinal = Vxi *time
solve for x final.Subtract 300+2500 and you have it. Check my thinking.
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