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.

First find the launch angle. See

http://www.jiskha.com/display.cgi?id=1298683188
for that.

Equations for range and maximum height can be found there.

Range does not depend upon mass.

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

First, let's find the maximum height attained by the projectile. In projectile motion, the time taken to reach maximum height is equal to the time taken to reach the same height when the projectile is coming down. This implies that the time of flight (total time taken) of the projectile is twice the time taken to reach maximum height.

The time taken to reach maximum height can be found using the formula:
time = (initial vertical velocity) / (acceleration due to gravity)

The initial vertical velocity can be calculated using the initial kinetic energy:
initial vertical velocity = √(2 * initial kinetic energy / mass of the projectile)

Now, we know that the range of the projectile (horizontal distance traveled) is given by:
range = (initial horizontal velocity) * (time of flight)

Since the range is equal to the maximum height attained, we can equate the above equation to the initial vertical velocity * time of flight.

Setting up the equation, we have:
range = (initial vertical velocity) * (time of flight)

Simplifying further, we have:
range = √(2 * initial kinetic energy / mass of the projectile) * (2 * time of flight)

We can rearrange this equation to solve for the range:
range = 2 * (√(2 * initial kinetic energy / mass of the projectile) * time of flight)

Now, let's substitute the given values into the equation and calculate the range:

Given:
Initial kinetic energy (K.E.) = 1036 J
Mass of the bullet = 0.1 kg
Acceleration due to gravity (g) = 9.81 m/s²

First, calculate the initial vertical velocity:
initial vertical velocity = √(2 * initial kinetic energy / mass of the projectile)
initial vertical velocity = √(2 * 1036 J / 0.1 kg)
initial vertical velocity ≈ 454.03 m/s

Next, calculate the time of flight:
time of flight = 2 * (initial vertical velocity) / (acceleration due to gravity)
time of flight = 2 * 454.03 m/s / 9.81 m/s²
time of flight ≈ 46.33 s

Finally, substitute the values into the range equation:
range = 2 * (√(2 * initial kinetic energy / mass of the projectile) * time of flight)
range = 2 * (√(2 * 1036 J / 0.1 kg) * 46.33 s)

Calculating the range:
range ≈ 454.03 m/s * 46.33 s
range ≈ 21063.34 m
range ≈ 21.06334 km

Therefore, the range of this projectile when fired at an angle such that the range equals the maximum height attained is approximately 21.06334 km.

To find the range of the projectile, we can use the equations of motion for projectile motion. The range (R) of a projectile is given by the equation:

R = (2 * v₀² * sinθ * cosθ) / g

Where v₀ is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

In this case, we are looking for the range when it equals the maximum height attained. At the maximum height, the vertical component of velocity is zero. Using this information, we know that the angle of projection (θ) is 45 degrees.

Now, let's solve for the initial velocity (v₀):

v₀² = (2 * K) / m

Where K is the initial kinetic energy and m is the mass of the bullet.

Plugging in the given values:

v₀² = (2 * 1036 J) / 0.1 kg
v₀² = 20690 J/kg

v₀ = √(20690 J/kg)
v₀ = 143.92 m/s

Now that we have the initial velocity, we can calculate the range when it equals the maximum height:

R = (2 * (143.92 m/s)² * sin45° * cos45°) / 9.81 m/s²

R = (2 * 20733.8 * 0.707 * 0.707) / 9.81
R = 203206.6 / 9.81
R = 20750.06 m

Finally, converting the range to kilometers:

R = 20750.06 m / 1000 m/km
R ≈ 20.8 km

Therefore, the range of the projectile when it is fired at an angle such that the range equals the maximum height attained is approximately 20.8 km.