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nelan · Sep 27, 2009, 12:02 PM

A rock is held steady over a cliff and dropped. 0.2 seconds later, another rock is thrown straight down at a speed of 12.1 m/s, and hits the first rock. How far have the rocks dropped before they collide? How long is the first rock in the air before it gets hits by the second rock?

Perito · Sep 27, 2009, 01:43 PM

The first rock travels this far

$D = \frac 12 GT^2$

where G is the acceleration due to gravity and D is the distance traveled and T is the time the rock is in the air.

The second rock uses a similar equation

$D = 12.2(T-0.2) + \frac 12 G(T-0.2)^2$

The two distances are equal. Set the two equations equal to each other and solve for T.

nelan · Sep 27, 2009, 05:51 PM

is g=-9.8 or g=9.8

Unknown008 · Sep 28, 2009, 08:34 AM

Think of it for a while. What sign did you assign the velocity of the two rocks?

If you assigned a positive sign to the downward direction, all the other motion in that direction will be positive, and the upward negative.

If you put them as negative, then the all of them in the downward direction will be negative and the upward positive.