Free fall and one other question

[wonger]

So i have two questions i need help answering. its not homework, but its from a previous exam my teacher gave.

x0=x knot and v0= v knot

Homework Statement

a rock dropped from a cliff falls one-third of its total distance to the ground in the last second of its fall. determine the height of the cliff

Homework Equations

v=v0+at
x=x0+v0+.5at^2
i don't think any of the other kinetic equations are relevant

The Attempt at a Solution

i know that the equation for the last second of freefall is x=x0+v0-4.9(1)^2 and v= v0-9.8.

im not sure where to start here. my first impression is that i have to find the final velocity before it hits the ground, but I am not sure what to do to find it.

second question

Homework Statement

a train pulls away from a station with a constant a cceleration of .4 m/s^2. a passenger arrives at the track 6.0s after the end of the rtain has passed the very same point. what is the slowest constant speed at which she can run and catch the train?

Homework Equations

v=v0+at
x=x0+v0+.5at^2
v^2=v0^2+2a(xf-x0)

The Attempt at a Solution

@6seconds, its position is 7.2m from starting point and its velocity is 2.4 m/s. I am not exactly sure what to do on this one either, i know that you have to "optimize" the problem by setting them equal to each other, but I am not really sure which equations to use cus any equation i use i get two variables.

 

[rl.bhat]

Hi wonger, welcome to PF.
Height of the cliff x = 0.5*g*t^2 ...(1)
Distance traveled in (t-1) s is
x1 = 0.5*g*(t-1)^2...(2)
In the problem it is given that
x - x1 = x/3. From these hints find t and then x.

 

[tomcue]

For the first question, you are correct that you need to find the final velocity before the rock hits the ground. To do this, you can use the equation v=v0+at, where v0 is the initial velocity (which is 0 since the rock is dropped), a is the acceleration due to gravity (which is -9.8 m/s^2), and t is the time it takes for the rock to fall. You can then use this final velocity in the equation x=x0+v0+.5at^2 to find the height of the cliff.

For the second question, you are on the right track with using the equations v=v0+at and x=x0+v0+.5at^2, but you also need to use the third equation v^2=v0^2+2a(xf-x0). This equation relates the initial and final velocities to the distance traveled. Since the passenger needs to catch the train at the same point where it was 6 seconds ago, you can set x0=0 and xf=7.2m. You can then use these three equations to solve for the initial velocity v0 and the time t it takes for the passenger to catch the train. Once you have these values, you can use the equation x=x0+v0t+.5at^2 to find the distance the passenger runs in order to catch the train. To find the slowest constant speed at which she can run, you can set this distance equal to the distance traveled by the train in 6 seconds (which is 7.2m). You can then solve for the initial velocity v0, which will give you the slowest speed at which she can run and still catch the train.