Physics Kinematics Homework Help ?

Click For Summary

Homework Help Overview

The problem involves kinematics, specifically the motion of two stones dropped from a height. The first stone is dropped from rest, and the second stone is thrown downward after a certain distance has been fallen by the first stone. The goal is to determine the initial velocity required for the second stone to reach the ground simultaneously with the first stone.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss calculating the time taken for the first stone to fall a specific distance and using that time to find the initial velocity of the second stone. There is mention of using kinematic equations to relate displacement, time, and acceleration.

Discussion Status

Some participants have provided calculations and proposed methods for solving the problem, while others question the correctness of the derived values. There is no explicit consensus on the final answer, but the discussion is exploring various approaches to the problem.

Contextual Notes

Participants note that the forum guidelines discourage providing complete solutions, emphasizing the importance of individual effort in homework tasks.

jayjay55531
Messages
3
Reaction score
0
Physics Kinematics Homework Help!?

"While standing on a bridge 18.3 m above the ground, you drop a stone from rest. When the stone has fallen 2.50 m, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction."

After finding the velocity of the first stone (AFTER falling 2.50 m) to be -7m/s, I tried to solve for initial velocity of the second stone by setting the times equal to each other (displacement for the first stone being 15.8, and for the second stone 18.3).

I used the equation x = v0*t + (1/2)*a*t2
and of course I got an ugly quadratic-equationey mess trying to get t on one side. There has to be an easier way to do this...
 
Physics news on Phys.org


I would suggest solving for t for the first stone since you know it's initial velocity is 0 and its acceleration is -g. Whatever you get for that time, plug it into the equation for the second stone. Then, solve the second equation for the initial velocity.
 


Hi, I tried to do this question as an exercise. Is the answer 2.06 m/s?

First, I find the time taken for the first stone to fall down 2.5m.

2.5 = (0)t + 1/2 (9.8) t^2
t= 0.71s

Then I find the total time for the stone to reach the ground.

18.3 = (0)t + 1/2 (9.8) t^2
t= 1.93s

The time left for the second stone to catch up is 1.93 - 0.71 = 1.22s

By using s = ut + 1/2 at^2,

18.3 = u(1.22) + 1/2 (9.8) (1.22)^2

u= 2.06m/s.

Correct?
 


coconut62 said:
Hi, I tried to do this question as an exercise. Is the answer 2.06 m/s?

First, I find the time taken for the first stone to fall down 2.5m.

2.5 = (0)t + 1/2 (9.8) t^2
t= 0.71s

Then I find the total time for the stone to reach the ground.

18.3 = (0)t + 1/2 (9.8) t^2
t= 1.93s

The time left for the second stone to catch up is 1.93 - 0.71 = 1.22s

By using s = ut + 1/2 at^2,

18.3 = u(1.22) + 1/2 (9.8) (1.22)^2

u= 2.06m/s.

Correct?

Please note that we do not do other students' work for them here at the PF. Normally your post would be deleted for showing the solution, but since it has been 2 weeks since the original post (OP), I'll leave it visible.
 

Similar threads

Replies
6
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 13 ·
Replies
13
Views
12K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
6
Views
2K
Replies
23
Views
4K
Replies
8
Views
2K