You throw a stone vertically upward with an initial speed of 6.1?

I'm going to take this a step farther for you. Set up the parametric equations as follows:

Horizontal: X(t) = Vo t cos w (where Vo = initial velocity, t = time, w = angle)

Vertical: Y(t) = Vo t sin w - g t^2 (where g = constant force of gravity in whatever units you're using)

In this example, w = pi/2, so X(t) = 0 and Y(t) reduces to Vo t - g t^2

Y = 6.1 t - g t^2. (Let's assume the units are meters, so g = 10 meters / second)

10 = 6.1 t - 10 t^2. Solve for t (it's just a quadratic equation)

Part B: Velocity = derivative of position. dY/dt = 6.1 - 20 t

1st solve for position = 0 = Y(t) = 6.1 t - 10 t^2. (Choose the non-trivial value of t = terminal time)

2nd: Input terminal t into dY/dt = 6.1 - 20t to calculate velocity. (Negative value indicates direction is downward towards the ground.)

The linear units are assumed in meteres .

Height reached from the window (H) = (6.1 * 6.1)/(2*9.8) = 1.89 m

Time of flight =

Time taken to reach Height H from the window + Time taken to fall through (H+10)

sqrt( 2H/g ) + sqrt ( 2(H+10) ) /g = sqrt (2* 1.89/9.8) + sqrt ( 2* 11.89/9.8) =0.621 + 1.55 =

2.17 sec

Velocity when it strikes ground = sqrt ( 2 * 9.8 * (H+10)) = sqrt ( 2* 9.8* 11.89) = 15.26 m/s