A rock is held steady over a cliff and dropped. 0.5 seconds later, another rock is thrown straight down at a speed of 9.9 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?
I'm not sure where to start.
I'm not sure where to start.
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Write down two equation for the rocks, and find where they meet.
The position of the first rock is given by
y = gt^2 / 2 where y is measured down form the top of the cliff.
the position of the second rock after 0.5 seconds is given by
y = 9.9(t - 0.5) + g(t--.5)^2 / 2
When they collide the y will be the same. therefore
gt^2 / 2 = 9.9(t - 0.5) + g(t--.5)^2 / 2
This equation gives you t, the answer to the second part. From this you can calculate y the distance the first rock dropped.
The position of the first rock is given by
y = gt^2 / 2 where y is measured down form the top of the cliff.
the position of the second rock after 0.5 seconds is given by
y = 9.9(t - 0.5) + g(t--.5)^2 / 2
When they collide the y will be the same. therefore
gt^2 / 2 = 9.9(t - 0.5) + g(t--.5)^2 / 2
This equation gives you t, the answer to the second part. From this you can calculate y the distance the first rock dropped.