A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.87 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?
S=ut-(1/2)gt²
S=41.7 m, u=0, g=9.8 m/s²
Solve for t.
Now
S=v0(t-1.87) - (1/2)g(t-1.87)²
Solve for v0.
To find the initial speed of the second stone, we can use the following formula for free-falling objects:
h = (1/2) * g * t^2 + v0 * t + h0
where:
- h is the height of the object above the water
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken
- v0 is the initial velocity of the object
- h0 is the initial height of the object
For the first stone, we have:
h1 = 41.7 m (height above the water)
g = 9.8 m/s^2
t = ? (unknown)
v01 = ? (unknown)
h01 = 0 m (initial height)
For the second stone, we have:
h2 = 0 m (height above the water)
g = 9.8 m/s^2
t = 1.87 s (time taken)
v02 = ? (unknown)
h02 = 41.7 m (initial height)
Since both stones strike the water at the same time, we can equate the times for the two stones, which gives us:
h1 = h2
(1/2) * g * t^2 + v01 * t + h01 = (1/2) * g * t^2 + v02 * t + h02
Substituting the known values and simplifying:
(1/2) * (9.8) * t1^2 + v01 * t1 + 0 = (1/2) * (9.8) * (1.87)^2 + v02 * (1.87) + 41.7
Simplifying further:
(4.9) * t1^2 + v01 * t1 = 8.6822 + 1.87 * v02 + 41.7
We are interested in determining the initial speed of the second stone, so let's solve for v02:
v02 = (4.9 * t1^2 + v01 * t1 - 8.6822) / 1.87 + 41.7
However, we still need the value of t1 and v01.
Given that the first stone is dropped (not thrown), its initial velocity, v01, is 0 m/s.
Therefore, the equation for v02 becomes:
v02 = (4.9 * t1^2 - 8.6822) / 1.87 + 41.7
To determine t1, we can use the equation:
h1 = (1/2) * g * t1^2
Solving for t1:
t1^2 = (2 * h1) / g
t1 = sqrt((2 * 41.7) / 9.8)
Now we can substitute this value of t1 back into the equation for v02 to find the initial speed of the second stone.