Kinmatics problem

  • Thread starter tony873004
  • Start date
  • #1
tony873004
Science Advisor
Gold Member
1,751
143
A ball, dropped from rest, covers 2/7 of the distance to the ground in the last two seconds of its fall.

(a) From what height was the ball dropped?
(b) What was the total time of fall?

If I can figure out either part a or part b, the other part will be easy.

I have a feeling that the solution to this will involve simultaneous equations. But I don't really know where to begin aside from making a list of everything I know:

[tex]d_{t} = d_{1} + d_{2}[/tex]
[tex]d_{t} = \frac {2}{7} d + \frac {5}{7} d[/tex]

[tex]t_{t} = t_{1} + t_{2} [/tex]
[tex]t_{t} = t_{1} + 2 seconds[/tex]

[tex]v_{i_{t}} = 0[/tex]
[tex]v_{f_{t}} = v_{f_{2}}[/tex]

[tex]d_{1} = \frac{5}{7}d_{t}[/tex]
[tex]t_{1} = t_{t} - t_{2}[/tex]
[tex]t_{1} = t_{t} - 2 seconds[/tex]

[tex]v_{i_{1}} = 0[/tex]
[tex]v_{f_{1}} = v_{i_{2}}[/tex]

[tex]d_{2} = \frac{2}{7} d_{t}[/tex]
[tex]t_{2} = 2 seconds[/tex]
[tex]v_{i_{2}} = v_{f_{1}}[/tex]
[tex]v_{f_{2}} = v_{f_{t}}[/tex]

Can anyone suggest a starting point?
 

Answers and Replies

  • #2
Pyrrhus
Homework Helper
2,178
1
Well, i will divide it in 2 parts, then i'll relate the equations with their link variable, the first part final speed, which will be the second part initial speed, Also i will use on the first part the equation without time, while on the second part i will use the displacement equation with time.
 
  • #3
1,789
4
Suppose at any time the y coordinate of the ball is y(t). Then the distance it has covered equals (h-y(t)). So,

[tex](h-y(t = T)) - (h-y(t = T-2)) = \frac{2h}{7}[/tex]

Additionally, y(t = T) = 0.

Cheers
vivek
 

Related Threads on Kinmatics problem

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
6
Views
710
Replies
1
Views
2K
Top