a major league outfielder throws a baseball horizontally with a velocity of 22 m/s, released from a height of 1.8m. how far does it go before it hits the ground ?

To find how far the baseball goes before hitting the ground, we can use the equations of motion. In this case, we're interested in the horizontal distance the ball travels.

Since the outfielder throws the ball horizontally, the initial vertical velocity is zero. We can use the equation:

h = (1/2) * g * t^2,

where h is the vertical displacement (1.8m in this case), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time the ball takes to reach the ground.

First, let's find the time it takes for the ball to hit the ground. We can use the formula:

t = sqrt((2 * h) / g),

where sqrt represents the square root function. Plugging in the values:

t = sqrt((2 * 1.8) / 9.8)
t ≈ sqrt(0.36735)
t ≈ 0.606 s

The time calculated is the total time of flight before the ball hits the ground.

Now, we can find the horizontal distance traveled by the ball using the equation:

d = v * t,

where d is the horizontal distance, v is the horizontal velocity (22 m/s in this case), and t is the time of flight.

Plugging in the values:

d = 22 * 0.606
d ≈ 13.27 m

Therefore, the baseball goes approximately 13.27 meters before hitting the ground.