Find the initial velocity of the second stone

[Josh chips]

Homework Statement

A stone is dropped from a point 320 meters above the ground; at the same instant another stone is thrown upward from a point 100 meters below the first. I f the two stones strike the ground simultaneously, what is the initial velocity of the second stone.

Relevant Equations

y=ut +1/2at^2

since the question states they hit the ground at the same time i firs find the time they hit the ground
y= ut + 1/2at^2 initial velocity is 0 for the first stone

320= 0 + 4.9t^2 this

t= 8 secs
since the 2nd stone was thrown 100 m below stone 1
distance from ground is 220m

220 = ut - 1/2at^2
220= 8u- 4.9*8^2
this give initial velocity as 66.7 m/s but am not sure if its the correct answer need some help

 

[DrClaude]

Why do you have a positive acceleration in the first case and a negative one in the second?

 

[DrClaude]

Note also that the equation you se doesn't take into account the position at $t=0$.

 

[Josh chips]

Why do you have a positive acceleration in the first case and a negative one in the second?

Why do you have a positive acceleration in the first case and a negative one in the second?

I have taken up as +ve and down as -ve since the first stone was dropped and the second stone was thrown upwards

 

[Josh chips]

Note also that the equation you se doesn't take into account the position at $t=0$.

At t=0 , we have the height from which it was thrown

 

[DrClaude]

I have taken up as +ve and down as -ve since the first stone was dropped and the second stone was thrown upwards

Your are not consistent in your choice of the positive direction. While it is ok to change convention between the first and the second part of the problem, this leads you to make a mistake in the second part.

So either redo the second part using the same convention as the first (I think this is the preferred method) or make sure that all the values in the second part are consistent with your new convention.

 

[Josh chips]

Here is how I have done it
For the second stone we have
-220 =ut -1/2at^2
-220 =8u -313.6
This gives u as 11.7 m/s
Kindly confirm if this is correct

 

[DrClaude]

The process is correct, but the rounding errors are quite important, due to truncating the time to 8 s. I get u > 12 m/s.

 

[neilparker62]

Might be worth noting that since the two stones have no relative acceleration you can simply write: $$ut = 100m$$ with t being the value you calculated in post #1 (just round correctly!).

 

[DrClaude]

Might be worth noting that since the two stones have no relative acceleration you can simply write: $$ut = 100m$$ with t being the value you calculated in post #1 (just round correctly!).

I advise strongly against that. When first dealing with such problems, do not take any short cuts. Work the problem from basic equations. As I wrote above, I am even bothered by the lack of an initial position in the equation used.

 

[neilparker62]

I advise strongly against that. When first dealing with such problems, do not take any short cuts. Work the problem from basic equations. As I wrote above, I am even bothered by the lack of an initial position in the equation used.

Ok - point concerning the use of basic equations accepted as being a more educationally correct method. Then one should rather begin by specifying ground level as reference and up as the positive direction and write:

$$ s_1=320-½gt^2...1$$ $$s_2=220+ut-½gt^2...2 $$ At ground level the stones have the same displacement from reference (zero in this case): $$ s_1-s_2 = 0 ⇒ 320-220-ut = 0 $$ And we note that in the subtraction the term $-½gt^2$ cancels out.