A baseball is thrown horizontally off a 43-m high cliff with a speed of 18.3 m/s. What will be the speed of the baseball in m/s just before it hits the ground?

See previous post.

To determine the speed of the baseball just before it hits the ground, we can use the principles of projectile motion. Since the baseball is thrown horizontally, it has no initial vertical velocity, only an initial horizontal velocity of 18.3 m/s.

First, we need to find the time it takes for the baseball to fall from a height of 43 m. We can use the equation:

h = (1/2)gt^2

where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. Plugging in the given values:

43 m = (1/2) * 9.8 m/s^2 * t^2

Simplifying the equation gives:

t^2 = (2 * 43 m) / 9.8 m/s^2
t^2 = 8.7755 s^2.

Taking the square root of both sides, we find that t ≈ 2.96 s.

Since there is no acceleration in the horizontal direction, the horizontal velocity remains constant throughout the motion. Therefore, the horizontal velocity just before hitting the ground remains 18.3 m/s.

Now, to find the speed just before hitting the ground, we can use the Pythagorean theorem:

v = √(v_horizontal^2 + v_vertical^2)

where v is the total speed, v_horizontal is the horizontal velocity, and v_vertical is the vertical velocity.

In this case, since the baseball was thrown horizontally, the vertical velocity is equal to zero. Thus, the equation simplifies to:

v = √(18.3 m/s)^2 + 0^2
v = √(334.89 m^2/s^2)
v ≈ 18.3 m/s.

Therefore, the speed of the baseball just before it hits the ground is approximately 18.3 m/s.