A stone is dropped into a river from a bridge 43.8 m above the water. Another stone is thrown vertically down 1.65 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?
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To find the initial speed of the second stone, we need to understand the concept of free fall and apply the equations of motion.
The first stone is dropped, which means it is in free fall. The only force acting on it is gravity, which causes it to accelerate downwards at a constant rate, known as the acceleration due to gravity (g).
Given:
Height (h) of the bridge = 43.8 m
Time taken for the first stone to strike the water (t1) = unknown
Time taken for the second stone to be thrown after the first stone (t2) = 1.65 s
We can use the equation of motion for free fall:
h = (1/2) * g * t^2
For the first stone:
Using the equation h = (1/2) * g * t1^2, we can rearrange it to find t1:
t1^2 = (2 * h1) / g
t1 = sqrt((2 * h1) / g)
Next, we need to find the time taken by the second stone (t2) to reach the water. Since it is thrown vertically downwards, it also undergoes free fall. However, it experiences an additional time delay of 1.65 seconds before it starts falling.
Therefore, t2 = t1 - 1.65
Given that both stones strike the water at the same time, t1 = t2.
So, t1 = t2 = t
Now, we can substitute t2 in terms of t:
t = sqrt((2 * h) / g) - 1.65
To find the initial speed (u) of the second stone, we can use the equation of motion:
h = ut + (1/2) * g * t^2
Since the final height (h) is zero (the water's surface), the equation becomes:
0 = ut + (1/2) * g * t^2
Simplifying, we get:
ut + (1/2) * g * t^2 = 0
Now, we can substitute the value of t = sqrt((2 * h) / g) - 1.65:
u * (sqrt((2 * h) / g) - 1.65) + (1/2) * g * (sqrt((2 * h) / g) - 1.65)^2 = 0
Solving this equation will give us the initial speed (u) of the second stone.