A person standing at the edge of a seaside cliff kicks a stone horizontally over the edge with a speed of 14.3 m/s. The cliff is 47.6 m above the water's surface. How far in meters from the base of the cliff does the stone land?

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To solve this problem, we can use the equations of motion for projectile motion. Since the stone is kicked horizontally, its initial vertical velocity is zero. The only force acting on the stone vertically is the force of gravity.

Here are the steps to find the horizontal distance:

Step 1: Determine the time taken for the stone to fall from the cliff to the water's surface. To find this, we can use the equation:

Δy = 1/2 * g * t^2

where Δy is the vertical displacement (the height of the cliff) and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Plugging in the values:
47.6 m = (1/2) * (9.8 m/s^2) * t^2

Rearranging the equation, we have:
t^2 = 47.6 m * 2 / 9.8 m/s^2

Solving for t, we get:
t = sqrt(47.6 m * 2 / 9.8 m/s^2)

Evaluating the expression, we find that t ≈ 3.07 seconds.

Step 2: Using the time calculated in the previous step, we can find the horizontal distance traveled by the stone. Since the stone is kicked horizontally with a constant speed, the distance can be found using the equation:

d = v * t

where d is the horizontal distance, v is the horizontal velocity of the stone, and t is the time of flight.

Plugging in the values:
d = 14.3 m/s * 3.07 s

Evaluating the expression, we find that d ≈ 43.8 meters.

Therefore, the stone lands approximately 43.8 meters from the base of the cliff.