A stone is dropped into a river from a bridge 45.4 m above the water. Another stone is thrown vertically down 1.14 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?
To find the initial speed of the second stone, we need to first calculate the time it takes for the first stone to reach the water.
We can use the equation for the displacement of an object undergoing free fall near the surface of the Earth:
s = ut + (1/2)gt^2
Where:
s = displacement (45.4 m)
u = initial velocity (unknown for the first stone)
t = time taken to reach the water (unknown for the first stone)
g = acceleration due to gravity (9.8 m/s^2)
Since the stone is dropped, its initial velocity u will be zero. Using this information, we can simplify the equation to:
s = (1/2)gt^2
Plugging in the values, we get:
45.4 = (1/2) * 9.8 * t^2
Simplifying further:
45.4 = 4.9t^2
Now, let's solve for t:
Divide both sides by 4.9:
9.28 = t^2
Taking the square root of both sides:
t ≈ 3.04 seconds
Now that we know it takes approximately 3.04 seconds for the first stone to reach the water, we can find the initial speed of the second stone.
Since the second stone is thrown vertically down, its initial velocity will be negative (-u) because it is moving in the opposite direction of the positive upward direction.
We can use the same equation for the displacement of an object undergoing free fall and set the displacement as -45.4 m (negative because it is moving downward):
-45.4 = -ut + (1/2)gt^2
Now, let's substitute the known values:
-45.4 = -u * 1.14 + (1/2) * 9.8 * (1.14)^2
Simplifying further:
-45.4 = -1.14u + 6.13476
Rearranging the equation:
-1.14u = -51.53476
Dividing both sides by -1.14:
u ≈ 45.25 m/s
Therefore, the initial speed of the second stone is approximately 45.25 m/s.