problem says that a cannon ball is fired at a 30 degree angle off a 10 meter high wall at 80 m/s. Use work and energy to find it's impact speed.

I worked out this out with the equation:
(1/2)(mass)(initial velocity)^2 + mgy = (1/2)(mass) (final velocity)^2 + mg(y final)
i got impact speed equals 81 m/s. I'm I right?? and if not, were did I go wrong?

initial speed: 80

added energy: mgh=10*9.8*m

total energy= 1/2 m 80^2+98 m
1/2 m vf^2=1/2 m 80^2+98m

vf=sqrt (80^2+98)=80.6m/s

I am not certain how many significant digits you had 10m,80m/s...

To solve this problem using work and energy, you need to consider the initial and final states of the cannonball.

Let's break it down step by step:

1. The initial state of the cannonball is at the top of the wall before it is fired. At this point, it has gravitational potential energy (mgy) due to its height above the ground (y = 10 m).

2. The final state of the cannonball is when it hits the ground. At this point, it has kinetic energy (1/2)(mass)(final velocity)^2 since it is in motion.

Now let's set up the equation using the principles of energy conservation:

Initial energy = Final energy

Gravitational potential energy (mgy) = Kinetic energy (1/2)(mass)(final velocity)^2

Substituting the given values and variables:

(mass)(gravitational acceleration)(10 m) = (1/2)(mass)(final velocity)^2

Since the mass cancels out on both sides of the equation, we can remove it:

(gravitational acceleration)(10 m) = (1/2)(final velocity)^2

Next, we need to find the gravitational acceleration (g). Assuming the cannonball is near the surface of the Earth, we can approximate it as 9.8 m/s².

(9.8 m/s²)(10 m) = (1/2)(final velocity)^2

98 m²/s² = (1/2)(final velocity)^2

98 m²/s² * 2 = (final velocity)^2

196 m²/s² = (final velocity)^2

Now, take the square root of both sides to obtain the final velocity:

final velocity = sqrt(196 m²/s²) = 14 m/s

Therefore, the impact speed of the cannonball is 14 m/s, not 81 m/s.