a gun converts 200J of stored energy into kinetic energy of the 0.02kg bullet. What is the speed of the bullet as it leaves the gun? If the gun is fired straight up, how high will the bullet go?

how high up?

200J=massbullet*g*heightinmeters

speed? 200=1/2 massbullet*velocity^2

To find the speed of the bullet as it leaves the gun, we can use the principle of conservation of energy. The total energy (E) in the system remains constant, so the initial potential energy (PE) is equal to the final kinetic energy (KE). We can calculate the initial potential energy using the given information:

PE = 200 J

The final kinetic energy can be calculated using the formula:

KE = (1/2)mv^2

Where:
m = mass of the bullet = 0.02 kg
v = velocity of the bullet

Since the bullet is fired, its initial velocity is 0. Thus, the final kinetic energy is equal to the total energy:

KE = E = 200 J

Now, let's substitute the values into the equation and solve for v:

200 J = (1/2)(0.02 kg)v^2

Multiplying both sides by 2 and dividing by 0.02 kg gives us:

v^2 = (200 J * 2) / 0.02 kg
v^2 = 20,000 J / 0.02 kg
v^2 = 1,000,000 m^2/s^2

Taking the square root of both sides, we find:

v ≈ 1000 m/s

Therefore, the speed of the bullet as it leaves the gun is approximately 1000 m/s.

To calculate how high the bullet will go when the gun is fired straight up, we can use the equation of motion:

(vf^2 - vi^2) = 2gh

Where:
vf = final velocity = 0 m/s (at the highest point)
vi = initial velocity = 1000 m/s (speed of the bullet leaving the gun)
g = acceleration due to gravity = 9.8 m/s^2
h = maximum height

Simplifying the equation, we get:

(0 - 1000^2) = 2 * 9.8 * h
-1000^2 = 19.6 * h

Solving for h:

h = (-1000^2) / (19.6)
h ≈ -510,204.1 m

The negative value indicates that the height is below the initial position of the bullet. Therefore, the bullet will not reach any significant height when fired straight up and will eventually fall back to the ground.