The initial kinetic energy imparted to a 0.1 kg bullet is 1036 J. The acceleration of gravity is 9.81 m/s2 . Neglecting air resistance, find the range of this projectile when it is fired at an angle such that the range equals the maximum height attained.

Answer in units of km.

To find the range of the projectile when it is fired at an angle such that the range equals the maximum height attained, we need to use the principles of projectile motion.

Let's break down the problem into two parts: finding the maximum height attained and finding the range.

1. Maximum Height Attained:
The maximum height is reached at the vertex of the projectile's trajectory, where the vertical component of its velocity becomes zero.
Using the equation for initial vertical velocity (v₀ₓ), time of flight (t), and acceleration due to gravity (g), we can find the maximum height (H):
H = (v₀ₓ²) / (2g)

2. Range of the Projectile:
The range of the projectile can be determined using the equation:
R = (v₀ₓ * t)
where v₀ₓ is the initial horizontal velocity of the projectile, and t is the total time of flight.

To determine the v₀ₓ, we can use the initial kinetic energy (KE) of the bullet:
KE = (1/2) * m * (v₀ₓ²)

Now, let's solve for the range R when the range equals the maximum height H.

First, find v₀ₓ:
KE = (1/2) * m * (v₀ₓ²)
1036 J = (1/2) * 0.1 kg * (v₀ₓ²)

Simplifying the equation:
v₀ₓ² = (2 * 1036 J) / 0.1 kg
v₀ₓ² = 2 * (1036 J / 0.1 kg)
v₀ₓ² = 2 * 10360 m²/s²
v₀ₓ = sqrt(2 * 10360) m/s

Now that we have v₀ₓ, let's find the time of flight (t):
t = 2 * (v₀ₓ / g)

Substituting the values:
t = 2 * (sqrt(2 * 10360) m/s / 9.81 m/s²)

Finally, we can find the range R by multiplying v₀ₓ with t and converting the answer to kilometers (km):
R = (v₀ₓ * t) / 1000 km

Let's plug in the values and calculate the range R in km.