While standing on a bridge 15.0 m above the ground, you drop a stone from rest. When the stone has fallen 3.10 m, you throw a second stone straight down. What 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
(-15.0)-(1/2)(-9.80)(.96)/.96 answer equals .10.7
Tina - I did this problem for Brittany, scroll down.
first get time to fall to ground
a = -9.8
v = -9.8 t
-15 = -4.9 t^2
t = sqrt (15/4.9) = 1.75 seconds
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Now how long did it take to reach 3.1 m down
3.1 = -4.9t^2
t = .795 seconds
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so the second rock spends 1.75 -.795 or .955 seconds in the air
-15 = Vo (.955) -4.9 (.955)^2
.955 Vo = 3.64-15
Vo = - 11.9
so 11.9 m/s downward (negative)
wrong answer -4.9(.955)^2 does not equal 3.64
To find the initial velocity needed for the second stone, we can use the kinematic equations of motion. The key is to recognize that both stones will reach the ground at the same time, meaning they will take the same amount of time to fall.
First, let's find the time it takes for the first stone to fall 3.10 m. We can use the kinematic equation:
y = y0 + v0t + (1/2)at^2
where:
- y is the final position (height) of the stone (3.10 m in this case),
- y0 is the initial position (height) of the stone (15.0 m in this case),
- v0 is the initial velocity of the stone (0 m/s since it is dropped from rest),
- t is the time taken to fall,
- a is the acceleration due to gravity (-9.8 m/s^2 in the downward direction).
Substituting the known values:
3.10 m = 15.0 m + 0 m/s * t + (1/2) * (-9.8 m/s^2) * t^2
Simplifying the equation:
-11.70 m = -4.9 m/s^2 * t^2
Rearranging the equation:
t^2 = (-11.70 m) / (-4.9 m/s^2)
t^2 = 2.39 s^2
Taking the square root of both sides:
t = √(2.39 s^2)
t ≈ 1.55 s
Now that we know the time taken for the first stone, we can find the initial velocity of the second stone using the same time but starting at a different height (0 m):
y = y0 + v0t + (1/2)at^2
0 m = 0 m/s * t + (1/2)(-9.8 m/s^2) * t^2
Simplifying the equation:
0 m = -4.9 m/s^2 * t^2
Rearranging the equation:
t^2 = 0 m / (-4.9 m/s^2)
t^2 = 0 s^2
Taking the square root of both sides:
t = √(0 s^2)
t = 0 s
From this equation, we can see that the second stone must be given an initial velocity in order to reach the ground at the same time as the first stone. Since time is the same, we can set up an equation for the height of the second stone using the time (1.55 s):
y = y0 + v0t + (1/2)at^2
0 m = 0 m/s * 1.55 s + (1/2)(-9.8 m/s^2)(1.55 s)^2 + 0
Simplifying the equation:
0 m = -4.9 m/s^2 * (2.4025 s^2)
Rearranging the equation:
v0 ≈ (0 m - (-4.9 m/s^2) * (2.4025 s^2)) / (1.55 s)
v0 ≈ (0 m + 11.94 m/s^2) / (1.55 s)
v0 ≈ 7.713 m/s
Therefore, in order for the second stone to reach the ground at the same time as the first stone, it must be given an initial velocity of approximately 7.713 m/s in the downward direction.