While standing on a bridge 45.0 m above ground, you drop a stone from rest. When the stone has fallen 3.80 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.

Question

While standing on a bridge 45.0 m above ground, you drop a stone from rest. When the stone has fallen 3.80 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.

Answer 1

While standing on a bridge 16.0 m above the ground, you drop a stone from rest. When the stone has fallen 3.80 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?

Answer 2

To solve this problem, we can use the equations of motion for free-falling objects. The first stone is dropped, so it falls freely. The second stone is thrown downward with an initial velocity.

Let's start by calculating the time it takes for the first stone to fall 3.80 m. We can use the equation:

h = (1/2) * g * t^2

Where h is the initial height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

Rearranging the equation, we get:

t = sqrt(2h / g)

Plugging in the values, we have:

t = sqrt(2 * 3.80 / 9.8)
t ≈ 0.785 s

Now, since both stones need to reach the ground at the same time, the second stone should take the same time to reach the ground, considering it was thrown from 3.80 m.

Next, let's find the time it takes for the second stone to fall from a height of 45.0 m. We'll use the same equation:

t = sqrt(2h / g)

Plugging in the values, we have:

t = sqrt(2 * 45.0 / 9.8)
t ≈ 3.02 s

Now, the time of flight for the second stone should also be around 0.785 s. Therefore, we can set up another equation using the equation of motion for the second stone.

Since the initial velocity of the second stone is unknown, we'll call it v₀.

The equation of motion for the second stone is:

h = v₀ * t + (1/2) * g * t^2

Substituting the values we know, we have:

0 = v₀ * 0.785 + (1/2) * 9.8 * 0.785^2

Simplifying the equation, we get:

0 = v₀ * 0.785 + 3.04

Solving for v₀, we find:

v₀ ≈ -3.04 / 0.785
v₀ ≈ -3.87 m/s

The initial velocity of the second stone should be approximately -3.87 m/s (negative because it is thrown downward) in order for both stones to reach the ground at the same instant.

aasa

While standing on a bridge 16.0 m above the ground, you drop a stone from rest. When the stone has fallen 3.80 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?

To solve this problem, we can use the equations of motion for free-falling objects. The first stone is dropped, so it falls freely. The second stone is thrown downward with an initial velocity.