Simple Physics problem

[Wnh5001]

"A stone is thrown vertically upward at a speed of 16m/s at time t=0. A second stone is thrown upward with the same speed 1 second later. At what time are the two stones at the same height?

hints. suggestions, and Exact equations are welcome.

 

[JDrake]

With no gravity, never

 

[feralkid]

Originally posted by: joedrake
With no gravity, never

Well there never is gravity; my school district calls it intelligent falling.

 

[sao123]

hint: the 2nd stone will peak exactly 1 second after the first.
So some time between when the 1st stone peaks, and when the second stone peaks is your answer.

 

[CollectiveUnconscious]

Before the stones were picked up to be thrown.

 

[Wnh5001]

Originally posted by: sao123
hint: the 2nd stone will peak exactly 1 second after the first.
So some time between when the 1st stone peaks, and when the second stone peaks is your answer.

yeah i figured it would between those two time intervals.. >_>, but hmm, i tried to solve for time for the first ball and got something but i think its wrong. i used Vf=Vo + at.

vf=0
-16m/s/9.81= 1.631 sec + 1 sec,

interval [1.631,2.631]?, but i need exact time.. and i think someone said to make up the equation for both, and set them equal to each other..

 

[drinkmorejava]

just set up the two delta X equations to equal eachother X = Vit + 1/2at^2 to solve for t and then use the equation again with the value of t to solve for x.

the eq for the second object will use (t + 1) for t

 

[Armitage]

Will the stones kill you when they come back down?
What if your standing on a conveyer that matches the speed of the plane (which does take off) when you throw the stones?

More seriosly...
Write the equation for the altitude of each stone from a common time point - like the time the first stone was thrown. Then set the equations equal to each other and solve for t.

 

[dullard]

We have to assume they were thrown from the same initial height.

We have obvious answer #1: at time t = -1 sec.
We have obvious answer #2: at time t -> infinity, assuming there are no other strange factors going on.

For the third time, just set the two equations equal to each other.

 

[blinky8225]

20.9/9.8 = ~2.133 seconds?

 

[JDrake]

Originally posted by: CollectiveUnconscious
Before the stones were picked up to be thrown.

I like that answer

 

[Wnh5001]

Originally posted by: blinky8225
20.9/9.8 = ~2.133 seconds?

correct, but what did u do to get that?

 

[sao123]

Originally posted by: drinkmorejava
just set up the two delta X equations to equal eachother X = Vit + 1/2at^2 to solve for t and then use the equation again with the value of t to solve for x.

the eq for the second object will use (t + 1) for t

ah... a helpful equation.

 

[QED]

It's been awhile since I sat in a physics classroom, but here it goes:

Speed of stone at time t is s(t) = 16 - gt

We integrate to get the height of stone at time t: h(t) = 16t - .5 * g * t^2

Hence, the height of stone 2 (thrown a second later) will be h2(t) = 16(t-1) - .5 * g * (t-1)^2.

Setting these two heights to be equal and solving:

16t - .5g * t^2 = 16(t-1) - .5g * (t-1)^2
16t - .5g * t^2 = 16t -16 - .5g * t^2 - .5g * 2 * t + .5g
gt = 16 + .5g,

or t=16/g + .5, or roughly 2.13 seconds.

 

[QED]

Originally posted by: drinkmorejava
just set up the two delta X equations to equal eachother X = Vit + 1/2at^2 to solve for t and then use the equation again with the value of t to solve for x.

the eq for the second object will use (t + 1) for t

Close, but you have to use (t-1) for t in the second equation...

 

[So]

This took more than :heart: minutes to solve! :|