physics with a falling ball

Chokfull · #1 $Old$ 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!

string6bean1977 · #2 $Old$ November 3rd, 2009, 01:35 PM

velocity can be negative, SPEED can not. speed is the abs val of V.

Chokfull · #3 $Old$ November 4th, 2009, 10:18 AM

Quote:

Originally Posted by string6bean1977 $View Post$

velocity can be negative, SPEED can not. speed is the abs val of V.

Yes, but I'm working with velocity, not speed. that's why it's a problem when it's negative; i worked using DOWN as positive, so if the velocity is negative at one point, that means the ball is flying.

skeeter · #4 $Old$ November 4th, 2009, 01:48 PM

if you dropped a ball from rest at the top of the window, it would take only about a half a second to pass the 1.2 m window on its way down. check it yourself ...

$t = \sqrt{\frac{2x}{a}}$

the problem says it takes 1.2 s to pass ... what does that tell you?

Chokfull · #5 $Old$ November 5th, 2009, 07:15 AM

You make a good point--if it was dropped from a higher point (the roof) then it must be going faster, and thus take less time to fall.

For it to take more time, it must be going slower, thus justifying the negative velocity.

Well, it may be a typo in the book. if so, it wasn't thee first

Paul46 · #6 $Old$ November 12th, 2009, 12:29 PM

Chokfull,

Hi i'm from PHF, i get 14.54m also.

V at bottom of window = (1.2/1.25) + ((1.25/2)*9.8) = 7.085 m/s

7.085/9.8 = 0.7229 secs

since the ball was dropped, we have s= (1/2)gt^2

s = 4.9*1.7229^2 = 14.54m