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A cannon ball leaves a fort with an initial horizontal speed of 180 m/s and strikes a ship in the sea below 7.2 s later. What is the ball's range? How high is the fort above sea level?

range = 7.2s * 180m/s = 1296 meters
height = .5 (9.8m/s2)(7.2s)2 ≈ 254.02 meters

First, time of the ball's flight is t=7.2s;

The ball's motion can be broken into two motions: horizontal, with constant velocity 180 m/s, and vertical, with initial velocity being zero and acceleration being equal to g=9.8 m/s2, due to gravity. The principle of superposition of motions allows to describe them independently.

The equation of motion is:
y=y0+vy0t+gyt2/2; Here vy0 and gy are projections of the initial velocity and acceleration of gravity onto the vertical axis, y0 is the initial position. Let the cannon position be the origin of our coordinate system. Then y0=0; We know that vy0=0, since the ball was shot horizontally. If the downward direction is positive, then gy=g=9.8. So we got for vertical motion:
y=gt2/2;
In 7.2 s y=9.8*7.22/2=254 m;

So the cannon (and the fort) is 254 m above the sea level.

The range of the ball is easy to find. Since the horizontal motion happens with constant velocity,
x=x0+vx0t;
Here x0 is initial position, vx0 is the projection of initial velocity onto horizontal axis. It is natural to choose the horizontal axis in the direction towards the ship, then vx0=v0, initial velocity. x0=0 since the origin coincides with the cannon. So
x=v0t; x=180*7.2=1296 m.

Answer: Range is 1296 m; fort is 254 m above the sea level.
Sun -- here's the "play-by-play" ... final Vdown is 72m/s ... ave Vdn is 36m/s ... height is 7.2*36 = 252 +7 or ~260m ... range is 7.2*180 or 7.2*200 with "10% discount" 1440-144 or 1296m :)