A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of 18m/s. The cliff is 52 meters above the waters surface. How long does it take for the stone to fall to the water? with what speed does it strike the water?
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24
36m/s
To find the time it takes for the stone to fall to the water, we can use the equation of motion for vertical motion under gravity:
Δy = v₀t + (1/2)gt²
where:
Δy = vertical displacement (52 meters in this case, as the stone is dropped from a 52 meter high cliff)
v₀ = initial velocity (18 m/s in this case)
t = time
g = acceleration due to gravity (approximately 9.8 m/s²)
Rearranging the equation to solve for t and plugging in the given values, we have:
52 = 18t + (1/2)(9.8)t²
Simplifying the equation, we get:
4.9t² + 18t - 52 = 0
Now we can solve this quadratic equation for t. We can use the quadratic formula:
t = (-b ± √(b² - 4ac))/(2a)
In this case, a = 4.9, b = 18, and c = -52. Plugging these values into the quadratic formula gives us:
t = (-18 ± √(18² - 4(4.9)(-52)))/(2(4.9))
Calculating further, we have:
t = (-18 ± √(324 + 1011.2))/(9.8)
Simplifying further, we get:
t ≈ (-18 ± √1335.2)/9.8
Now we can calculate the two possible roots of the equation and obtain the positive value because time cannot be negative:
t₁ ≈ (-18 + √1335.2)/9.8 ≈ 2.93 seconds
t₂ ≈ (-18 - √1335.2)/9.8 ≈ -6.06 seconds
Hence, the stone takes approximately 2.93 seconds to fall to the water.
Now, to find the speed at which the stone strikes the water, we can use the equation:
v = v₀ + gt
Plugging in the values, we get:
v = 18 + (9.8)(2.93)
Calculating further, we have:
v ≈ 45.16 m/s
Therefore, the stone strikes the water with a speed of approximately 45.16 m/s.