free fall

• Aug 31st 2009, 03:19 PM
BiGpO6790
free fall
At time t = 0 s, apple 1 is dropped from a bridge onto a roadway beneath the bridge; somewhat later, apple 2 is thrown down from the same height. Figure 2-25 gives the vertical positions y of the apples versus t during the falling, until both apples have hit the roadway. With approximately what speed is apple 2 thrown down?
http://edugen.wiley.com/edugen/cours...2/fig02_25.gif
• Aug 31st 2009, 07:59 PM
CaptainBlack
Quote:

Originally Posted by BiGpO6790
[IMG]file:///C:/Users/Nick/AppData/Local/Temp/moz-screenshot-3.jpg[/IMG]At time t = 0 s, apple 1 is dropped from a bridge onto a roadway beneath the bridge; somewhat later, apple 2 is thrown down from the same height. Figure 2-25 gives the vertical positions y of the apples versus t during the falling, until both apples have hit the roadway. With approximately what speed is apple 2 thrown down?
http://edugen.wiley.com/edugen/cours...2/fig02_25.gif

Take the origin to be the release point, and \$\displaystyle y\$ positive downwards.

The time it takes the first apple to reach the roadway tells you how high the roadway is. Since for this the distance from the release point is:

\$\displaystyle y=gt^2/2\$

and we know that \$\displaystyle t=2\$ seconds when \$\displaystyle y\$ is the bridge height.

For the second apple we know that the time to hit the road is \$\displaystyle 1.25\$ seconds. The apples distance from the release point is:

\$\displaystyle y=gt^2/2 + v_0t\$

where t is the time from release of the second apple.

As we know the height of the bridge we set this to \$\displaystyle y\$, and the time of fall which we set to \$\displaystyle t\$ we can now solve for the speed with which the apple is thrown down \$\displaystyle v_0\$

CB