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?
d1 = 0.5g*t^2 = 4.9*1.14^2 = 6.37 m. head start.
h = 0.5g*t^2 = 45.4 m.
4.9t^2 = 45.4
t^2 = 9.27
Tf = 3.04 s. = Fall time of both stones
d2 = d1+ 6.37 m.
Vo*t + 0.5g*t^2 = 0.5g*t^2 + 6.37
Vo*t + 0.5g*t^2 -0.5g*t^2 = 6.37
Vo*t = 6.37
Vo*3.04 = 6.37
Vo = 2.09 m/s.
NOTE: The 2 Free-fall Eqs cancel.
Correction:
d1=0.5g*t^2=4.9*1.14^2=6.37 m Head start.
V = Vo + g*t = 0 + 9.8*1.14=11.17m/s. =
Velocity after falling 6.37 m.
h1 = Vo*t + 0.5g.t^2 =45.4-6.37
11.17t + 4.9t^2 = 39
4.9t^2 + 11.17t - 39 = 0
Tf = 1.9 s. = Fall time of both stones.
h2 = Vo*t + 0.5g*t^2 = 45.4
1.9Vo + 4.9*1.9^2 = 45.4
1.9Vo = 45.4-17.7 = 27.7
Vo = 14.6 m/s. = Initial velocity of 2nd
stone.
To find the initial speed of the second stone, we need to work through the problem step-by-step. Let's start by determining the time it took for the first stone to hit the water.
Given:
- Height of the bridge (h) = 45.4 m
First, we need to calculate the time it took for the stone to hit the water when dropped from the bridge.
Using the equation: h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity (which is approximately 9.8 m/s^2), and t is the time.
Rearranging the equation, we get: t = √(2h/g)
Substituting the values, we have:
t1 = √(2 x 45.4 / 9.8)
Calculating this, t1 ≈ 3.02 seconds (rounded to two decimal places).
Now, we need to determine the time it took for the second stone to hit the water after being thrown vertically down. We are given that the second stone was thrown vertically down 1.14 seconds after the first stone was dropped.
Therefore, the time taken by the second stone can be calculated as follows:
t2 = t1 - 1.14
Substituting the known value of t1, we have:
t2 = 3.02 - 1.14
Calculating this, t2 ≈ 1.88 seconds (rounded to two decimal places).
Since both stones hit the water at the same time, we can assume that the time taken is the same for both stones. Therefore, we can equate t1 and t2.
t1 = t2
Simplifying the equation, we have:
√(2 x 45.4 / 9.8) = t2
Taking the square of both sides, we have:
2 x 45.4 / 9.8 = t2^2
Calculating this value, we find:
t2 ≈ 1.57 seconds (rounded to two decimal places).
Now, we can find the initial speed of the second stone using the equation: v = at, where v is the initial velocity, a is acceleration, and t is time.
Since the stone is thrown vertically downward, the acceleration due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction).
Substituting the known values, we have:
v = -9.8 x 1.57
Calculating this, v ≈ -15.37 m/s (rounded to two decimal places).
Note that the negative sign indicates that the velocity is directed downward.
Hence, the initial speed of the second stone is approximately 15.37 m/s (rounded to two decimal places).