A student throws a set of keys vertically upward to his fraternity brother, who is in a window 3.95 m above. The brother's outstretched hand catches the keys 1.80 s later. (a) With what initial velocity were the keys thrown?

d = Vo*t + 4.9t^2 = 3.95m,

Vo*1.8 + (-4.9)(1.8)^2 = 3.95,
1.8Vo - 15.876 = 3.95,
1.8Vo = 3.95 + 15.876 = 19.826,
Vo = 11m/s.

To determine the initial velocity of the keys, we can use the kinematic equation for vertical motion. The equation relating the displacement (Δy), initial velocity (v0), time (t), and acceleration due to gravity (g) is:

Δy = v0t - (1/2)gt^2

In this case, the displacement (Δy) is the height of the window above the student's hand, which is 3.95 m. The time (t) is 1.80 s, and the acceleration due to gravity (g) is approximately 9.8 m/s^2.

Substituting these values into the equation, we have:

3.95 = v0(1.80) - (1/2)(9.8)(1.80)^2

Now we can solve this equation for v0.

First, simplify the equation:

3.95 = 1.8v0 - 8.82

Next, rearrange the equation to isolate v0:

1.8v0 = 3.95 + 8.82

1.8v0 = 12.77

Finally, divide both sides by 1.8 to solve for v0:

v0 = 12.77 / 1.8

v0 ≈ 7.094 m/s

Therefore, the initial velocity of the keys when thrown upward was approximately 7.094 m/s.