Lots of Trouble With Free-Fall

  • Thread starter evilempire
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  • #1
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Hi, it's my first post. Thanks for letting me join your forums as I learn physics.

Here is the problem:

A stone is thrown vertically upward at a speed of 35.30 m/s at time t=0. A second stone is thrown upward with the same speed 1.390 seconds later. At what time are the two stones at the same height?

I've used kinematics to get the time for the stone to reach maximum height as 3.6 seconds, for what that's worth. I've been trying to figure this one out for a while, and am effectively lost. Anything to help me get in the right direction would be extremely appreciated.

Thank you for reading.
 

Answers and Replies

  • #2
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You know these equations (I hope at least)

Yfinal = A/2t^2 + Vit + Yi
Vfinal = At + Vi
Where A = acceleration
Vi = initial velocity
Yi = initial height

So…

Yfinal = -4.9t^2 + 35.3t + 0 is going to be the position equations for when a rock is first thrown and Vfinal = -9.8t + 0 will be your velocity equation.

So I would first figure out the velocity and y position of the first rock thrown (rock A) when you throw the second rock (rock B)

RockA’s Y = 4.9(1.390)^2 + 35.3(1.390) + 0
Since I don’t have a calculator on me let us call this value N
RockA’s Vfinal = -9.8(1.390) + 0
And we will call this one M

So the equation of the first rock when the second one is thrown is:
RockA’s Yfinal equation When rockB is launched = -4.9t^2 + M*t + N

Your goal is to find out when the two Yfinals are the same, so you set the equations equal to each other…
-4.9t^2 + 35.3t = -4.9t^2 + M*t + N
and solve for t
 
  • #3
24
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JonF said:
You know these equations (I hope at least)

Yfinal = A/2t^2 + Vit + Yi
Vfinal = At + Vi
Where A = acceleration
Vi = initial velocity
Yi = initial height

So…

Yfinal = -4.9t^2 + 35.3t + 0 is going to be the position equations for when a rock is first thrown and Vfinal = -9.8t + 0 will be your velocity equation.

So I would first figure out the velocity and y position of the first rock thrown (rock A) when you throw the second rock (rock B)

RockA’s Y = 4.9(1.390)^2 + 35.3(1.390) + 0
Since I don’t have a calculator on me let us call this value N
RockA’s Vfinal = -9.8(1.390) + 0
And we will call this one M

So the equation of the first rock when the second one is thrown is:
RockA’s Yfinal equation When rockB is launched = -4.9t^2 + M*t + N

Your goal is to find out when the two Yfinals are the same, so you set the equations equal to each other…
-4.9t^2 + 35.3t = -4.9t^2 + M*t + N
and solve for t
Thanks for the detailed and thorough response! I believe I now understand the concept of the problem, but am unsure how to solve that particular equation for t. I understand the -4.9t^2 will cancel each other out, but how does one get the t's on one side? Please explain and thank you again.
 
  • #4
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If anyone else could jump in and help me out it would be great.
 
  • #5
17
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That's kind of basic, though I must admit I only got good at it recently. Just subtract the M*t from one side and add it to the other, with opposite signs. Always remember you may apply anything to any side of the equation, as long as you do it to the other side, too. So subtract it from one of the sides, you must subtract it from the other and you get opposite signs.
Good luck!
 

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