**1296 meters**

^{2})(7.2s)

^{2}≈

**254.02 meters**

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?

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range = 7.2s * 180m/s = **1296 meters**

height = .5 (9.8m/s^{2})(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/s^{2}, due to gravity. The principle of superposition of motions allows to describe them independently.

Start with the vertical one.

The equation of motion is:

y=y_{0}+v_{y0}t+g_{y}t^{2}/2; Here v_{y0} and g_{y} are projections of the initial velocity and acceleration of gravity onto the vertical axis, y_{0} is the initial position. Let the cannon position be the origin of our coordinate system. Then y_{0}=0; We know that v_{y0}=0, since the ball was shot horizontally. If the downward direction is positive, then g_{y}=g=9.8. So we got for vertical motion:

y=gt^{2}/2;

In 7.2 s y=9.8*7.2^{2}/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=x_{0}+v_{x0}t;_{
}

Here x_{0} is initial position, v_{x0} is the projection of initial velocity onto horizontal axis. It is natural to choose the horizontal axis in the direction towards the ship, then v_{x0}=v_{0}, initial velocity. x_{0}=0 since the origin coincides with the cannon. So

x=v_{0}t; x=180*7.2=1296 m.

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