So a ball is dropped from a roof, passing a window 1.2 m tall. It takes 1.25s to pass the said window. It takes 1 more second below the window before it hits the ground. How tall is the building?

$\displaystyle

x=vt+\frac {1} {2}at^2$, where $\displaystyle v$ is the starting velocity, $\displaystyle x$ is the distance, $\displaystyle t$ is the time, and $\displaystyle a$ is the acceleration.

Therefore, $\displaystyle v=\frac {x-\frac {1} {2}at^2} {t}$, or

$\displaystyle v=\frac {1.2-\frac {1} {2}*9.8*1.25^2} {1.25}$ for the starting velocity when it passes the top of the window.But this results in -5.165, which isn't possible unless down is taken to be negative, which it isn't.

I completed the problem without realizing this, only substituting a variable,z, for this number.

Another formula gave me the time taken to travel the distance above the window: $\displaystyle z=v+at$, where, this time, v is the starting speed when it was dropped (0). $\displaystyle t=\frac {z} {a}=\frac {z} {9.8}$

Then we add all the times together to find the total fall time, $\displaystyle \frac {z} {9.8}+1.25+1$, which, for simplicity, we will callb.

Then the last formula we use is $\displaystyle x=vt+.5at^2$, which i think is the same one we used earlier. this gives us $\displaystyle x=(0)(b)+(.5)(9.8)(b)^2$

$\displaystyle (.5)(9.8)(\frac {-5.165} {9.8}+2.25)^2=14.54$

but it's supposed to be 20.4

thanks to anyone who helps!