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?
To determine how far the stone lands from the base of the cliff, we need to calculate the horizontal distance the stone travels before hitting the water's surface.
First, we need to find the time it takes for the stone to hit the water. We can use the equation:
y = y0 + v0y * t + 0.5 * a * t^2
Where:
- y is the height of the cliff above the water's surface (47.6 m),
- y0 is the initial height of the stone (0 m),
- v0y is the vertical component of the stone's velocity (0 m/s),
- a is the acceleration due to gravity (-9.8 m/s^2),
- t is the time.
Since the stone is kicked horizontally, the initial vertical velocity (v0y) is 0 m/s. Therefore, the equation simplifies to:
y = y0 + 0 * t + 0.5 * a * t^2
47.6 = 0 + 0.5 * (-9.8) * t^2
Now we can solve for t:
47.6 = -4.9 * t^2
Dividing both sides by -4.9:
t^2 = 9.7142857142857143
Taking the square root:
t ≈ 3.12 s
Now that we know the time it takes for the stone to hit the water, we can calculate the horizontal distance using the equation:
x = v0x * t
Where:
- x is the horizontal distance,
- v0x is the horizontal component of the stone's velocity (14.3 m/s),
- t is the time.
Since the stone is kicked horizontally, the initial horizontal velocity (v0x) remains constant at 14.3 m/s. Therefore, the equation simplifies to:
x = 14.3 * 3.12
x ≈ 44.6 m
Therefore, the stone lands approximately 44.6 meters from the base of the cliff.