Kinematics Problems

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Homework Statement


An object is dropped from a tower. If it had been thrown down at 60m/s, it would have taken half the time. What is the height of the tower? (acceleration is 10m/s^ down)


Homework Equations


d= v1(t)+1/2a(t)^ ----> I think you could use this

a=v2-v1/t --> maybe


The Attempt at a Solution


a= -10m/s^
v=60m/s
t=?
d=?
t=v/a
t=60/-10
t=-6 s (which doesn't make sense since time can't be negative in this case)

If it helps the answer is 320 m. I'd really appreciate it if you could help me and explain step by step! :) thanks
 

Answers and Replies

  • #2
When the object is dropped, the initial velocity is zero.
So the equation becomes ( if you take down as negative )
-d = 0 +-1/2*g*t^2 ...(1)
In the second case you can write
-d = -vt/2 - 1/2*g*t^2/4...(2) Solve the two equations to find t. From that you can find d.
 
  • #3
^ Why did you make "d" negative?

I also don't understand why you have two formulas :( and what I'm doing in each and what I'm trying to find in each!
 
Last edited:
  • #4
^ Why did you make "d" negative?

I also don't understand why you have two formulas :( and what I'm doing in each and what I'm trying to find in each!

There are 2 formulas because each describes one of the situations. Given the time relationship given in the problem you use t/2 for the second equation.

You need two equations because you have two unknowns. You don't know h and you don't know t.
 
  • #5
^oh, thank you! :D
 
  • #6
Hello! I have another question ..sorry!

Problem


Mr. Jones lives 50 km away from you. You both leave home at 5:00 and drive toward the highway. Mr.Jones travels at 35km/h and you drive at 40km/h. At what time will you pass Mr.Jones?


From the book the answer says: 5:40

This is what i did.. i think I'm on the right track but I have no idea what to do next...

You

d= x
v= 40km/h
t =?

Mr.Jones
d= x+50
v= 35 km/h
t= ?

:eek::eek:
 
  • #7
Since Mr.Jones is ahead of you, when you meet him he travels x km and you travel x + 50 km. Since both of you start simultaneously, time of travel is same.
So time T = distance / velocity.
Find the time taken by each and equate it. From that you can get x. Then you can find T.
 
  • #8
Since Mr.Jones is ahead of you, when you meet him he travels x km and you travel x + 50 km. Since both of you start simultaneously, time of travel is same.
So time T = distance / velocity.
Find the time taken by each and equate it. From that you can get x. Then you can find T.

sorry i made an error the question is at what time will you pass ms.jones on the highway?

Does it change anything?
 
  • #9
okay when i equated it i got

40t=x+50 and 35t=x

now what do i do with the equations?
 
  • #10
okay when i equated it i got

40t=x+50 and 35t=x

now what do i do with the equations?

(x + 50)/40 = x/35.
Solve for x. the put it in one of the above equations to get time.
 
  • #11
(x + 50)/40 = x/35.
Solve for x. the put it in one of the above equations to get time.



so i basically cross-multiplied and got 35(x+50) = 40x
which then gave me 35x + 1750 = 40x
1750= 40x-35x
1750=5x
350=x

sub x=350 into x/35

350/35 = 10s

That doesn't make sense because the final time is supposed to be 5:40 (basically 40 minutes to add onto 5:00)
 
  • #13
Well, your answer should be 10 hours, not 10 seconds. Second, unless there's something we're missing, the answer your book gives doesn't make any sense because if Mr. Jones lives 50 km away, it will take you over an hour just to make up that distance.
 
  • #14
Unless the cars are following something other than a one dimensional path or the data you've given us is wrong, that answer is incorrect!
 
  • #15
Unless the cars are following something other than a one dimensional path or the data you've given us is wrong, that answer is incorrect!

No, the answer's perfectly fine because two people are driving towards each other, and the question asks when they'll meet. That means neither of them has to make it to the other person's home.
 
  • #16
Why would you be driving in opposite directions if you're driving towards the same highway? Unless the highway is in between the two houses of course but that's not clear from the problem.
 

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