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
To find the initial velocity required for the second stone to reach the ground at the same time as the first stone, we can use the equations of motion.
Given:
Initial height (h) = 15.0 m
Vertical displacement of the first stone (s1) = 3.10 m
Vertical acceleration due to gravity (g) = -9.8 m/s^2 (negative because it acts in the downward direction)
Using the equation of motion for vertical displacement:
s = ut + (1/2)at^2
For the first stone, substituting the known values:
3.10 = 0*t + (1/2)*(-9.8)*t^2
Rearranging the equation:
4.9t^2 - 3.10 = 0
Now, let's solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 4.9, b = 0, and c = -3.10.
t = (-0 ± √((0)^2 - 4*(4.9)*(-3.10))) / (2*4.9)
t = ± √(0 - (-60.4)) / 9.8
t = ± √(60.4) / 9.8
As we are considering the positive time for both stones, t = √(60.4) / 9.8
Now, we know that the time taken for the second stone to reach the ground will also be √(60.4) / 9.8, since we want them to reach the ground at the same time.
Let's assume the initial velocity of the second stone as v.
Using the equation of motion for vertical displacement again:
s = ut + (1/2)at^2
For the second stone, substituting the known values:
15.0 = v*(√(60.4) / 9.8) + (1/2)*(-9.8)*(√(60.4) / 9.8)^2
15.0 = v*(√(60.4) / 9.8) + (1/2)*(-9.8)*(60.4 / 9.8)
15.0 = v*(√(60.4) / 9.8) - (1/2)*60.4
Rearranging the equation:
v*(√60.4 / 9.8) = 15.0 + (1/2)*60.4
v*(√60.4 / 9.8) = 15.0 + 30.2
v*(√60.4 / 9.8) = 45.2
v = 45.2 / (√60.4 / 9.8)
v ≈ 13.99 m/s
Therefore, in order for the second stone to reach the ground at the same time as the first stone, it must have an initial velocity of approximately 13.99 m/s in the downward direction.
To solve this problem, we'll use the equations of motion for free fall. We'll start by finding the time it takes for the first stone to reach a height of 3.10 m.
1. Use the equation for the final position in free fall:
s = ut + (1/2)gt^2,
where s is the displacement, u is the initial velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s^2).
Since the stone is dropped from rest, the initial velocity of the first stone (u1) is 0, and the displacement (s1) is 3.10 m. We'll solve this equation for t1:
3.10 = 0*t1 + (1/2)(-9.8)(t1)^2
3.10 = -4.9(t1)^2
(t1)^2 = -3.10/-4.9
(t1)^2 = 0.63265
t1 ≈ 0.795 seconds.
Next, we'll find the total time it takes for the second stone to reach the ground. Since the first stone has already fallen 3.10 m, the distance for the second stone to fall is 15.0 m - 3.10 m = 11.9 m.
2. Use the same equation for the total distance (s2) with the initial velocity (u2) as the unknown:
s2 = u2t2 + (1/2)gt2^2
The displacement (s2) is 11.9 m, and the time (t2) is the same as t1 since we want both stones to reach the ground at the same time. We can substitute the known values and solve for u2:
11.9 = u2(0.795) + (1/2)(-9.8)(0.795)^2
11.9 = 0.795u2 - 3.919
15.819 ≈ 0.795u2
u2 ≈ 19.9 m/s (rounded to one decimal place).
Therefore, in order for both stones to reach the ground at the same instant, the second stone must be given an initial velocity of approximately 19.9 m/s, directed downwards.
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)