Please help me with this physics problem?

Question

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?

Answer 1

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.

 

Start with the vertical one.

 

The equation of motion is:

y=y₀+v_y0t+g_yt²/2; Here v_y0 and g_y 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 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=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+v_x0t;

Here x₀ 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=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.

Answer 2

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 :)

 

 

 

Answer 3

 

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

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