While standing on a bridge 18.5 m above the ground, you drop a stone from rest. When the stone has fallen 2.80 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.
d = 0 + 0.5*9.8t^2 = 2.8m.
4.9t^2 = 2.8,
t^2 = 2.8 / 4.9 = 0.57,
t = 0.76s to drop 2.8m.
d = 0.5 * 9.8t^2 = 18.5m,
4.9t^2 = 18.5,
t^2 = 18.5 / 4.9 = 3.78,
t1 = sqrt3.78 = 1.94s. = Time in flight
for 1st stone.
t2 = 1.94 - 0.76 = 1.18s. = Time in flight for 2nd stone.
d=Vi*(1.18) + 0.5*9.8*(1.18)^2=18.5,
1.18Vi + 6.82 = 18,5,
1.18Vi = 18.5 - 6.82 = 11.68,
Vi = 11.68 / 1.18 = 9.9m/s.
To determine the minimal initial velocity required for the second stone to reach the ground at the same time as the first stone, we can use the equation of motion for both stones.
For the first stone, we can consider the equations of motion for constant acceleration, since only gravity is acting on it. The equation for the displacement of an object under constant acceleration is:
y = y0 + v0t + (1/2)gt^2
Where:
y = displacement of the stone at time t
y0 = initial displacement of the stone
v0 = initial velocity of the stone
g = acceleration due to gravity
t = time
For the first stone, it starts from rest, so its initial velocity (v0) is 0. The initial displacement (y0) is 18.5 m, and the displacement (y) when it has fallen 2.80 m is -2.80 m. Using these values, we can rearrange the equation to solve for time (t):
y = y0 + v0t + (1/2)gt^2
-2.80 = 18.5 + 0t + (1/2)(-9.8)t^2
Simplifying the equation, we get:
-9.8t^2 = 21.3
Now we can solve for the time (t). Rearranging the equation, we have:
t^2 = -21.3 / -9.8
t^2 = 2.173
Taking the square root of both sides, we find:
t = √2.173
t ≈ 1.47 seconds
Now, we need to find the minimal initial velocity (v0) for the second stone, so it reaches the ground at the same time as the first stone. Since both stones continue to fall independently from the same height (18.5 m) when the second stone is thrown, the equation for the second stone's motion is:
y = y0 + v0t + (1/2)gt^2
Where:
y = displacement of the stone at time t
y0 = initial displacement of the stone
v0 = initial velocity of the stone
g = acceleration due to gravity
t = time
However, in this case, the initial displacement (y0) is 0, as the second stone is thrown from the same height as the first stone starts. We want to find the velocity (v0) when the displacement (y) is -18.5 m (the height from which both stones started). Rearranging the equation, we get:
y = y0 + v0t + (1/2)gt^2
-18.5 = 0 + v0(1.47) + (1/2)(-9.8)(1.47)^2
Simplifying the equation, we have:
-18.5 = 1.47v0 - 6.786
1.47v0 = -18.5 + 6.786
1.47v0 = -11.714
Now we can solve for the initial velocity (v0):
v0 = -11.714 / 1.47
v0 ≈ -7.97 m/s (rounded to two decimal places)
So, the minimal initial velocity you must give the second stone is approximately -7.97 m/s (in the downward direction), to reach the ground at the same instant as the first stone.