Acannon tilted up at a 33.0° angle fires a cannon ball at 74.0 m/s from atop a 18.0 m -high fortress wall. what is the ball's impact speed on the ground below?

The cannon ball hits the ground with a speed of 76.35 m/s.

Explanation:

We shall use the conservation of energy principle to solve the problem

The initial energy of the cannon ball is the sum of the potential and the kinetic energies.

We know that potential energy =

Applying the given values we obtain the initial potential energy as

Similarly the initial kinetic energy of the ball equals

Applying values we get

Thus the initial energy is thus

Now upon hitting the ground the only energy that the cannon ball will posses is kinetic energy since the potential energy of any object upon touching the surface of earth equals zero

Thus we have

Equating initial and final energies we get

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14