Ball Thrown up in air, Velocity Question

AI Thread Summary
A ball is thrown upward at 11 m/s, followed by a stone thrown at 25 m/s one second later. The equations of motion for both objects are derived to find when the stone catches up to the ball, leading to a calculated time of approximately 0.256 seconds after the stone is thrown. To determine their velocities at the same height, the velocity equation is used, factoring in the time difference. The discussion emphasizes the importance of aligning the time variables and suggests using graphical methods for verification. The final conclusion is that the time for the stone to catch the ball is indeed 0.256 seconds, considering the one-second delay in its launch.
cbrowne
Messages
30
Reaction score
0

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??
 
Physics news on Phys.org
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.
 
Back
Top