Ball Thrown up in air, Velocity Question

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Homework Help Overview

The problem involves two objects, a ball and a stone, thrown vertically upward with different initial speeds and at different times. The original poster seeks to determine the time it takes for the stone to catch up with the ball and their respective velocities at that point.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss setting up equations for the heights of both the ball and the stone over time, considering their initial velocities and the effects of gravity. Some participants suggest evaluating the equations at specific time intervals to simplify the problem.

Discussion Status

There is an ongoing exploration of the problem, with participants checking calculations and discussing the implications of their findings. Some guidance has been offered regarding the approach to take, particularly in evaluating the equations at the moment the stone is thrown.

Contextual Notes

Participants question whether to adjust the time for the stone's delayed launch and discuss the implications of this on their calculations. There is also mention of visualizing the trajectories to better understand the problem.

cbrowne
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Homework Statement



A ball is thrown vertically upward with an initial speed of 11 m/s. One second
later, a stone is thrown vertically upward with an initial speed of 25 m/s. (a) Find
the time it takes the stone to catch up with the ball. (b) Find the velocities of the
stone and the ball when they are at the same height.

Homework Equations





The Attempt at a Solution



a = 9.8
Vi,ball= 11 m/s Vi,stone= 25m/s

y= y0+ V0t+ 1/2 at2

y = 11t+ 1/2 (9.8) t2

y= 11t - 4.9 t2 ======= ball

y = 25 (t+1) + (4.9 (t+1))2

y = 25t + 25 - 4.9 ( t2 + 2t + 1)

y = 25 t + 25 - 4.9 t2-9.8t - 4.9

y= -4.9t2 + 15.2t + 20.1 ===== stone

11t - 4.9 t2 = -4.9t2 + 15.2t + 20.1



am i on the right track??
 
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Sort of.

I think it might be easier to take a snap shot at the start of the second launching.

At that moment the ball is

Yo = 11*t - ½gt² evaluated at 1 sec

And at 1 sec, the velocity V1 = 11 - 9.8*t evaluated at 1 sec.

Now you can write your equations all with the same time can't you?

The ball now can be given by:

Y = (11 - ½g) + (11-9.8)*t - ½gt²

The rock:

Y = 0 + 25*t - ½gt²

When they are equal:

25*t - ½gt² = (11-4.9) + 1.2*t - ½gt²

And it looks like the ½gt² disappears.
 
so i get t=0.256 ... is this right? so this is the time it takes for the stone to catch up to the ball. how could i figure out B) the velocity?
 
You can check your answer for reasonability by drawing a graph. Do you know what kind of curve these trajectories make in the y,t plane? Draw both of them on the same graph (with the proper shift), and you should see a particular point of interest.
 
sorry, i have absolutely no idea how to draw that lol
 
cbrowne said:
sorry, i have absolutely no idea how to draw that lol
Hmm. What do you mean by "this"? Of course, the first step would be to decide what is the shape. A mathematical equation y=f(t) for each of the two balls would suffice. Then, you can either use the (antiquated, but useful on tests) skill of sketching this out by hand, or you can use Excel (or some other spreadsheet/plotting program).
 
cbrowne said:
so i get t=0.256 ... is this right? so this is the time it takes for the stone to catch up to the ball. how could i figure out B) the velocity?

What is the equation for the rock velocity?

V = Vo - a*t

You know t, so ...
 
should i add 1 second to my time?
 
cbrowne said:
should i add 1 second to my time?

You would add 1 sec if you wanted to know how long the ball was in the air.
 
  • #10
o ok.. i think i understand now.. so my answer is then i guess t= 0.256, since it was thrown 1 second after.
 
  • #11
This was originally why I suggested you consider taking the snapshot at the moment the rock was thrown and solving for t, because that's the answer you are looking for.
 

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