Solve Kinematics Problem: Dropped Object from Tower Height

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Homework Help Overview

The discussion revolves around a kinematics problem involving an object dropped from a tower and a related scenario involving two individuals driving towards each other. The original poster presents a situation where an object is dropped and questions the height of the tower based on time comparisons with a thrown object. The second problem involves calculating the time two individuals will meet while traveling towards a highway.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of kinematic equations to relate distance, velocity, and time for the dropped object scenario. There is confusion regarding the sign convention for distance and the necessity of using two equations for the two different situations. In the driving problem, participants discuss setting up equations based on relative distances and speeds, with some questioning the interpretation of the problem and the feasibility of the given answer.

Discussion Status

The discussion is active, with participants providing insights and clarifications on the kinematic equations and the setup of the driving problem. Some participants express confusion about the assumptions made in the problems, particularly regarding the negative sign for distance and the interpretation of the driving scenario. There is no clear consensus on the correctness of the answers provided, particularly in the driving problem.

Contextual Notes

Participants note that the original problem states specific conditions, such as the acceleration due to gravity and the initial velocities involved. There is also mention of potential misunderstandings regarding the setup of the driving problem, including the distances and directions of travel.

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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
 
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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.
 
^ 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:
Smile101 said:
^ 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.
 
^oh, thank you! :D
 
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:
 
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.
 
rl.bhat said:
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?
 
okay when i equated it i got

40t=x+50 and 35t=x

now what do i do with the equations?
 
  • #10
Smile101 said:
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
rl.bhat said:
(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)
 
  • #12
:frown:?
 
  • #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
jgens said:
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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