A 3 kg ball is thrown horizontally off a 78.5 m cliff with a speed of 12 m/s. How long does the ball stay in the air?

To find out how long the ball stays in the air, we can use the equations of motion for projectiles.

Step 1: Determine the initial vertical velocity (Vy).
Since the ball is thrown horizontally, there is no initial vertical velocity. Therefore, Vy = 0 m/s.

Step 2: Determine the vertical displacement (Δy).
The ball is thrown off a 78.5 m cliff, so the vertical displacement is Δy = -78.5 m (negative because the ball is moving downwards).

Step 3: Determine the acceleration in the vertical direction (ay).
The only force acting on the ball in the vertical direction is gravity. The acceleration due to gravity is approximately 9.8 m/s². Therefore, ay = -9.8 m/s² (negative because gravity acts downward).

Step 4: Use the kinematic equation to find the time of flight (t) or the total time the ball stays in the air.
The equation we'll use is: Δy = Vy*t + (1/2)*ay*t²

Plugging in the known values:
-78.5 = 0*t + (1/2)*(-9.8)*t²

Simplifying the equation:
-78.5 = -4.9t²

Divide both sides by -4.9 to isolate t²:
t² = 78.5 / 4.9

Taking the square root of both sides to solve for t:
t = √(78.5 / 4.9)

Calculating:
t ≈ 4.06 seconds

Therefore, the ball stays in the air for approximately 4.06 seconds.