Question

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?

Answer 1

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

Answer 2

Answer:

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