a flower pot is theown out of a window with a horizontal velocity of 8 m/s. If the window is 1.5 m off the ground, how far from the window does it land?

T = The time it takes for the flower pot to pass the top of my window.

V = The velocity of the flower pot at the moment it is passing the top of my window.

X = The height above the top of my window that the flower pot was dropped.

h = Lw + X

Lw = (1/2) * g*t^2 + V*t

V*t = Lw - (1/2) * g*t^2

V = Lw/t - (1/2) * g*t, On the other hand we know : V=gT.

Therefore we will have: Tg = Lw/t - (1/2) * g*t

T = Lw / (tg) - t/2

Now substitute for T in the following equation: X = (1/2) * g*T^2

X = (1/2) * g * (Lw / (tg) - t/2) ^2

Now substitute for X in the very first equation I mentioned: h = Lw + X

h = Lw + (1/2) * g * (Lw / (tg) - t/2) ^2

In case you wanted the answer to be simplified, then:

h = (Lw^2) / (2*g*t^2) + (g*t^2) / 8 + Lw/2

3.26m

Explanation:

Using one of the equation of motion to get the distance of the pot from the window and the ground;

v² = u²+2as where

v is the final velocity = 8m/s

u is the initial velocity = 0m/s

a = + g = acceleration due to gravity (this acceleration is positive since the body is falling downwards)

g = 9.81m/s

s is the distance between the object and the window from which it dropped.

Substituting this values to get the distance s we have;

8² = 0²+2 (9.81) s

64 = 19.62s

s = 64/19.62

S = 3.26m