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 determine how long the ball stays in the air, we can solve for the time it takes for the ball to reach the ground.
To start, let's analyze the given information:
- Initial velocity in the horizontal direction (vx) is 12 m/s.
- The ball is thrown horizontally, so there is no initial vertical velocity (vy = 0).
- The acceleration due to gravity is acting vertically downward (g = 9.8 m/s^2).
- The height of the cliff (h) is 78.5 m.
First, we can find how long it takes for the ball to reach the ground. In the horizontal direction, there is no acceleration (ax = 0), so the horizontal velocity remains constant throughout the motion.
Since we want to find the time the ball stays in the air, we can use the equation:
time = distance / speed
The distance the ball moves horizontally before hitting the ground is equal to the horizontal velocity multiplied by the time of flight (time). In this case, the distance is 78.5 m (the width of the cliff). Therefore:
78.5 m = 12 m/s * time
Now we can solve for time:
time = 78.5 m / 12 m/s
time ≈ 6.54 seconds
So, the ball stays in the air for approximately 6.54 seconds when thrown horizontally off the cliff.