A ball is thrown from the window of a high-rise building with an initial velocity of 6.1 m/s at an angle of 32 deg. below the horizontal. The bill strikes the ground 4.4 seconds after being thrown. Find the height from which the ball was thrown.

To find the height from which the ball was thrown, we can use the equations of motion for projectile motion.

First, we need to resolve the initial velocity into its horizontal and vertical components.

The horizontal component of the initial velocity is given by:
Vx = V * cos(theta)
where Vx is the horizontal component of the velocity, V is the magnitude of the velocity (6.1 m/s), and theta is the angle below the horizontal (32 degrees).

Vx = 6.1 * cos(32)
Vx ≈ 5.16 m/s

The vertical component of the initial velocity is given by:
Vy = V * sin(theta)
where Vy is the vertical component of the velocity.

Vy = 6.1 * sin(32)
Vy ≈ 3.29 m/s

Now, we can use the equation for vertical displacement (assuming the initial height is zero):
y = Vy * t + (1/2) * g * t^2
where y is the displacement in the vertical direction (height), t is the time (4.4 seconds), and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Plugging in the values:
y = 3.29 * 4.4 + (1/2) * 9.8 * (4.4)^2
y ≈ 60.58 meters

Therefore, the height from which the ball was thrown is approximately 60.58 meters.