A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of vi = 11 m/s. The cliff is h = 47 m above the water's surface, as shown below.
You gave the details but failed to ask the question.
If you want to know how far away from the (vertical) cliff the stone hits the water, multiply 11 m/s by the time T that it takes the stone to fall to the water's surface.
(g/2)T^2 = 47 m
T = sqrt (2*47/9.8)= 3.1 s
To analyze the situation, we can use the equations of motion in physics.
First, let's determine the time it takes for the stone to hit the water's surface. We can use the equation of motion for vertical motion under constant acceleration:
h = vi * t + (1/2) * g * t^2
Where:
h = height (47 m)
vi = initial velocity (11 m/s)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Rearranging the equation to solve for t, we get:
(1/2) * g * t^2 + vi * t - h = 0
Since this is a quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = (1/2) * g, b = vi, and c = -h.
Plugging in the values, we have:
t = (-vi ± √(vi^2 - 4 * (1/2) * g * (-h))) / (2 * (1/2) * g)
Simplifying further, we get:
t = (-vi ± √(vi^2 + 2gh)) / g
Now we can calculate the time it takes for the stone to hit the water's surface.
t = (-11 ± √(11^2 + 2 * 9.8 * 47)) / 9.8
Solving this equation gives us two possible solutions, one positive and one negative. Since time cannot be negative in this context, we take the positive solution:
t = (-11 + √(11^2 + 2 * 9.8 * 47)) / 9.8
Calculating this, we get:
t ≈ 3.41 seconds
Therefore, it takes approximately 3.41 seconds for the stone to hit the water's surface.