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Chokfull · #1 November 3rd, 2009, 01:20 PM

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?

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

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

$v=\frac {1.2-\frac {1} {2}*9.8*1.25^2} {1.25}$ 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: $z=v+at$ , where, this time, v is the starting speed when it was dropped (0). $t=\frac {z} {a}=\frac {z} {9.8}$

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

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

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

but it's supposed to be 20.4

thanks to anyone who helps!