A stone is thrown directly upward from rest from ground level with initial speed 20.0

m/s; 1.50 s after that, a second stone is thrown directly upward from ground level with initial
speed 18.0 m/s. (a) When are the stones at the same height? (b) What are the velocities of the
stones when they meet, in m/s?

To answer these questions, we need to understand the motion equations for objects in free fall.

Let's start by analyzing the motion of the first stone. It is thrown directly upward, so its initial velocity (u) is 20.0 m/s, and it experiences the acceleration due to gravity (g) in the downward direction. The upward direction is taken as positive, so the acceleration due to gravity is -9.8 m/s^2.

Using the equation for displacement in vertical motion:

h = ut + 1/2 * g * t^2

where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time.

For the first stone, at t = 1.50 s, substituting the given values:

h1 = (20.0 m/s) * (1.50 s) + 0.5 * (-9.8 m/s^2) * (1.50 s)^2

Now let's analyze the motion of the second stone, which is thrown after 1.5 seconds. This means its time (t) starts from 0 at this point. The initial velocity (u) of the second stone is 18.0 m/s.

h2 = (18.0 m/s) * t + 0.5 * (-9.8 m/s^2) * t^2

To find the time when the two stones are at the same height (h1 = h2), we can equate the two equations above and solve for t:

(20.0 m/s) * (1.50 s) + 0.5 * (-9.8 m/s^2) * (1.50 s)^2 = (18.0 m/s) * t + 0.5 * (-9.8 m/s^2) * t^2

Simplifying and rearranging the equation will give us the value of t when the stones are at the same height.

Once we have the time (t), we can substitute it back into either of the two height equations to find the height at that time, which will be the height where the stones meet.

To find the velocities of the stones when they meet, we need to differentiate the height equation with respect to time (t) to obtain the velocity equation:

v = u + g * t

where v is the velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time.

Using this equation, we can substitute the time (t) when the stones meet to find their velocities.

Now, let's solve these equations to get the answers.

To solve this problem, we can use the kinematic equations of motion to determine the height and velocity of each stone at any given time.

Let's start by determining the height and velocity of the first stone at time t. We can use the equation:

h = h0 + v0t - (1/2)gt^2

where:
h = height at time t
h0 = initial height (ground level in this case, so h0 = 0)
v0 = initial velocity
g = acceleration due to gravity (approximated as 9.8 m/s^2 for simplicity)
t = time

For the first stone, with an initial speed of 20.0 m/s, we have:

h1 = 0 + 20.0t - (1/2)(9.8)t^2

Now, let's determine the height and velocity of the second stone at time t. Using the same equation:

h2 = 0 + 18.0t - (1/2)(9.8)t^2

(a) To find when the stones are at the same height, we can set h1 = h2 and solve for t:

20.0t - (1/2)(9.8)t^2 = 18.0t - (1/2)(9.8)t^2
20.0t = 18.0t
2.0t = 0
t = 0

Therefore, the stones are at the same height initially when t = 0.

(b) To find the velocities of the stones when they meet, we can differentiate the position equations with respect to time:

v1 = v0 - gt
v2 = v0 - gt

For the first stone, at t = 0:

v1 = 20.0 m/s - (9.8 m/s^2)(0)
v1 = 20.0 m/s

For the second stone, at t = 0:

v2 = 18.0 m/s - (9.8 m/s^2)(0)
v2 = 18.0 m/s

Therefore, the velocities of the stones when they meet are 20.0 m/s and 18.0 m/s for the first and second stones, respectively.