While standing on a bridge 18.3 m above the ground, you drop a stone from rest. When the stone has fallen 2.60 m, you throw a second stone straight down. What minimal initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction.
Hey, that is the same as the one I just did for you, just different numbers.
To solve this problem, we can use basic kinematic equations and the principle of conservation of energy.
Let's start by analyzing the motion of the first stone. It falls from a height of 18.3 m, so we can use the kinematic equation:
d = vit + (1/2)at^2
Where:
- d is the distance traveled (2.6 m in this case)
- vi is the initial velocity (0 m/s since the stone is dropped from rest)
- a is the acceleration (which we assume to be the acceleration due to gravity, -9.8 m/s^2)
- t is the time
Rearranging the equation to solve for time t:
t = sqrt((2d)/a)
Substituting the given values, we find:
t = sqrt((2 * 2.6) / (-9.8)) ≈ 0.64 s (approximately)
Now, since we want the second stone to reach the ground at the same instant, we know that the time it takes for the second stone to fall must also be 0.64 seconds.
Next, let's analyze the motion of the second stone. It falls from a height of 18.3 m, so we can use the kinematic equation again:
d = vit + (1/2)at^2
Since the final position of the second stone is the same as the first stone (both reach the ground), the distance traveled is 18.3 m. The initial velocity (vi) is what we need to find.
Substituting the values into the equation, we have:
18.3 = vi * 0.64 + (1/2)(-9.8)(0.64)^2
Rearranging the equation to solve for vi:
vi = (18.3 - 0.5 * (-9.8) * (0.64)^2) / 0.64
Calculating this expression, we find that the minimal initial velocity vi of the second stone should be approximately 4.52 m/s (rounding to two decimal places).
Therefore, to ensure that both stones reach the ground at the same instant, you must give the second stone a minimal initial velocity of 4.52 m/s downward.