While standing on a bridge 45.0 m above ground, you drop a stone from rest. When the stone has fallen 3.90 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.
To solve this problem, we can use the equations of motion to find the initial velocity of the second stone.
Let's consider the motion of the first stone. We know that the initial position (s1) of the first stone is 45.0 m above the ground, and the distance it has fallen (Δs1) is 3.90 m.
The equation for the position of an object in free fall is given by:
s = ut + (1/2)gt^2
where:
s is the position,
u is the initial velocity,
t is the time, and
g is the acceleration due to gravity.
Since the stone is dropped from rest, the initial velocity (u1) of the first stone is 0 m/s.
We can rearrange the equation for the first stone to find the time it takes to fall the given distance:
3.90 = 0 + (1/2)(9.8)t^2
7.80 = 4.9t^2
t^2 = 7.80 / 4.9
t^2 = 1.59
t ≈ 1.26 seconds
Now, for the second stone to reach the ground at the same time as the first stone, it needs to have the same time of flight (t) and reach the ground from a height of 45.0 m. The equation for the position of the second stone is:
s = ut + (1/2)gt^2
where:
s is the position (which is 0 m, as it reaches the ground),
u is the initial velocity (what we want to find),
t is the time (1.26 seconds), and
g is the acceleration due to gravity (-9.8 m/s^2).
Plugging in the values into the equation:
0 = u2(1.26) + (1/2)(-9.8)(1.26)^2
Simplifying the equation gives:
0 = u2 * 1.26 - 7.6746
To solve for u2, we isolate it:
u2 * 1.26 = 7.6746
u2 = 7.6746 / 1.26
u2 ≈ 6.1048 m/s
Therefore, to reach the ground at the same instant as the first stone, the second stone must be initially thrown with an approximate velocity of 6.1048 m/s downwards.