# Find the initial velocity of the second stone

• Josh chips
In summary: So we have: $$s_1=320-½gt^2...1$$ $$s_2=220+ut-½gt^2...2$$This gives the following equations for the two stones:s_1=320-½gt^2...1s_2=220+ut-½gt^2...2
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

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

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

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

DrClaude said:
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

Josh chips said:
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.

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

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.

Josh chips
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!).

Last edited:
neilparker62 said:
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.

DrClaude said:
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.

DrClaude

## What is the meaning of "Find the initial velocity of the second stone"?

This phrase refers to determining the starting speed of the second stone in a given scenario, usually in a physics or math problem.

## How is the initial velocity of the second stone calculated?

The initial velocity of the second stone is typically calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time elapsed.

## What information is needed to find the initial velocity of the second stone?

To find the initial velocity of the second stone, you will need to know the final velocity, acceleration, and time elapsed. Additional information such as the distance traveled or the angle of launch may also be required depending on the specific problem.

## Can the initial velocity of the second stone be negative?

Yes, the initial velocity of the second stone can be negative. This indicates that the stone is moving in the opposite direction of the positive axis. However, in some problems, the initial velocity may be assumed to be positive unless otherwise specified.

## Why is it important to determine the initial velocity of the second stone?

The initial velocity of the second stone is important because it helps us understand the motion and behavior of the stone. It can also be used to predict the future path of the stone and make accurate calculations for other variables such as time and distance.

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