# Up and Down Motion (Two objects meeting in time?)

PROBLEM:

Suppose than an object is thrown upward with an initial velocity of 200ft/sec and that another one is thrown upward 5 seconds later with an initial velocity of 300ft/sec. When and where do they meet?

CONCERNS/QUESTIONS:

The answer I arrive at seems correct from my own understanding of the problem. However, I do not understand what this problem is teaching me? What is the relationship among these two objects when they meet at the same time (t) with the same displacement (s)? Does it mean they have the same slope "velocity"? I am confused on how I should attempt to solve this problem.

MY STRATEGY:

I found the expression for both vectors, (a) acceleration and (v) velocity, for both objects by using integration. I then found the expression for displacement (s) for both objects using integration.

Object 1 $$s=-16t^2+200t\\ \vec{v}=-32t+200\\\vec{a}=-32$$

Object 2 $$s=-16t^2+460t\\\vec{v}=-32t+460\\\vec{a}=-32$$

Then from here I was lost in what to do. Therefore, I just found $t$ for both objects when there slope is equal to $0$. This gave me the time at each objects maximum height.

Object 1
$t=6.25$ seconds

Object 2
$t=14.375$ seconds

I divided Object 2's (t) value by Object 1's (t) value and got $2.3$ seconds. This seems correct in my own mind after thinking long and hard about it.

$14.375 / 6.25$ = $2.3$ seconds after the second object has been thrown

I do not understand these last few steps or if I was even right. I'm not understanding the concept from which the problem is trying to teach me. May someone address my concerns and questions I have mentioned.

## 2 Answers

Two objects meet if you find points on their respective trajectories that have the same location and the same time.

Given what I expect you mean by all of your variables, you need to solve the equation

$$s_1 = s_2$$

where I've added the subscripts you did not. e.g. $s_1$ is the position of the first object.

And your formula for $s_2$ is incorrect: you forgot to account for the constant of integration, which needs to be carefully chosen to match the one known position of the second object. (i.e. it's on the ground when it gets thrown up)

It is a mistake to reuse $s,v,a$ for the two objects. I will use a subscript $_2$ for the second one. Your expressions for object 1 are fine. For object 2, the initial velocity is $300$, not $460$, but your equation for the velocity produces the correct result for $t \ge 5$. It might be more informative to present it as $v_2=-32(t-5)+300$ When you integrate to find $s_2$, you need to supply another constant of integration so that $s_2(5)=0$. A more intuitive way to give it would be $s_2=-16(t-5)^2+300(t-5)$ for $t \ge 5$ Now solve $s_1(t)=s_2(t)$ because that is when they meet.

Added: Setting $$-16(t-5)^2+300(t-5)=-16t^2+200t\\-16t^2+160t-400+300t-1500=-16t^2+200t\\ 460t-1900=200t\\t=\frac{1900}{260} \approx 7.3077$$

• When I set these two equations equal to each other and solve for (t) I get 5 seconds. This doesn't seem correct. Am I missing a step? Commented May 2, 2013 at 22:51
• Disregard, I had to graph it to see it. 2.3+5=7.3seconds Thanks for the correction on integration. I misinterpreted the procedure in determining the constant. Commented May 2, 2013 at 23:24
• @ShaneYost: I think that is a coincidence. You should not be looking for the time of maximum height and certainly should not be dividing their times. Your result is unitless as it is the ratio of two times. We want a result with units of time, the time when $s_1=s_2$ Commented May 2, 2013 at 23:27