A stone is dropped into a river from a bridge 43.3 m above the water. Another stone is thrown vertically down 1.06 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?
h=gt²/2 =>
t= sqrt(2h/g) = sqrt(2•43.3/9.8) =2.97 s.
t1=t-1.06 = 2.97 – 1.06 = 1.91 s.
h= v₀•t1 +g•t1²/2
Solve for v₀.
To find the initial speed of the second stone, we can first calculate the time it takes for the first stone to reach the water. Since the first stone is dropped from rest, we can use the equation for free fall:
s = ut + (1/2)gt^2
where:
- s is the distance fallen (in this case, 43.3 m)
- u is the initial velocity (0 m/s for a dropped stone)
- t is the time taken (which we want to find)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Rearranging the equation to solve for t, we get:
t = √(2s/g)
Substituting the given values, we have:
t = √(2 * 43.3 m / 9.8 m/s^2)
t ≈ √(8.806)
t ≈ 2.968 s
So, it takes approximately 2.968 seconds for the first stone to reach the water.
Since the second stone is thrown vertically down 1.06 seconds after the first is dropped, we can calculate the time it takes for the second stone to reach the water by subtracting the throwing delay from the time it took for the first stone to reach the water.
Time taken for the second stone = Time taken for the first stone - Throwing delay
= 2.968 s - 1.06 s
= 1.908 s
Now, we can use the equation for motion under constant acceleration to find the initial velocity of the second stone:
v = u + at
where:
- v is the final velocity (0 m/s when the stone reaches the water)
- u is the initial velocity (which we want to find)
- a is the acceleration (equal to the acceleration due to gravity, -9.8 m/s^2, since the stone is moving downward)
- t is the time taken (1.908 s)
Rearranging the equation to solve for u, we have:
u = v - at
= 0 m/s - (-9.8 m/s^2 * 1.908 s)
= 18.7184 m/s
So, the initial speed of the second stone is approximately 18.7184 m/s.