Your friend is standing below the window outside of our classroom and attempting to throw a rock past the window to get your attention. The rock is thrown straight up in the air and passes our window 0.40s after being released. It passes the same window on its way back down 1.60s later. What was the initial velocity of the rock? Did the height of the

Question

Your friend is standing below the window outside of our classroom and attempting to throw a rock past the window to get your attention. The rock is thrown straight up in the air and passes our window 0.40s after being released. It passes the same window on its way back down 1.60s later. What was the initial velocity of the rock? Did the height of the

window matter?

Answer 1

once past the window, the rock spent .8 seconds going up and .8 seconds going down.

So, the rock spent 1.2 seconds going up.

since v = Vo-32t and v(1.2)=0,

Vo-32(1.2)=0
Vo = 38.4

Doesn't matter how high the windows is, though its location is determined by the time intervals indicated.

Answer 2

To determine the initial velocity of the rock, we can use the kinematic equation for vertical motion:

h = vt + (1/2)gt^2

Where:
h = height (of the window, in this case)
v = initial velocity
t = time taken
g = acceleration due to gravity (approximately 9.8 m/s^2)

Since the rock passes the window twice, once on the way up and once on the way down, we know that the time taken for the entire journey is twice the time taken for it to pass the window once.

Let's calculate the time taken for the rock to pass the window on the way up and on the way down separately:

Time taken on the way up: 0.40 seconds
Time taken on the way down: 1.60 seconds

Now, we can use the equation for the motion of the rock on the way up to determine the initial velocity. Since the final velocity at the top of the path is 0 m/s, we have:

0 = v - gt

Solving for v gives:

v = gt

Next, we can use the equation for the motion of the rock on the way down to determine the height of the window. Since the rock takes 1.60 seconds to pass the window on the way down, we have:

h = vt + (1/2)gt^2

Substituting the value of v we found earlier, the equation becomes:

h = (gt)(1.60) + (1/2)g(1.60)^2

Simplifying further, we have:

h = 1.6gt + 1.6^2(g/2)

Now, we have two equations:

v = gt
h = 1.6gt + 1.6^2(g/2)

We need to solve for v and h to find the answer.

The height of the window does matter because it determines the value of h, which affects the calculations. If the height of the window is not provided, we cannot determine the initial velocity of the rock accurately.