Free fall and one other question

  • Thread starter Thread starter wonger
  • Start date Start date
  • Tags Tags
    Fall Free fall
AI Thread Summary
A rock dropped from a cliff falls one-third of its total distance in the last second of its fall, prompting a calculation of the cliff's height using kinematic equations. The relevant equations include v = v0 + at and x = x0 + v0 + 0.5at^2. The second question involves a train accelerating at 0.4 m/s², with a passenger arriving 6 seconds later, requiring the determination of the minimum speed needed to catch the train. The position of the train after 6 seconds is calculated to be 7.2 meters, with a velocity of 2.4 m/s. The discussion emphasizes optimizing the equations to solve for the variables involved in both scenarios.
wonger
Messages
1
Reaction score
0
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.
 
Last edited:
Physics news on Phys.org
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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top