# Stone dropping - yet another free fall acceleration problem

• missrikku
In summary, the two stones are thrown from the same bridge and hit the water at the same time. The first stone is dropped and takes 2.99 seconds to reach the water. The second stone is thrown 1 second after the first and takes 1.99 seconds to reach the water. Using the equations for motion, the initial velocity of the second stone can be calculated to be 12.31 m/s. When solving problems like these, it is important to carefully consider the given information and use the appropriate equations to find the solution.
missrikku
I'm sorry, I am just having difficulty with being sure of myself when it comes to these problems. This one states:

A stone is dropped into a river from a bridge 43.9m above the water. Another stone is thrown vertically down 1.00 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?

Well, I did this:

Y-Yo = 43.9m
a = -9.8 m/s^2

Vo1 = 0 m/s because it was just dropped

Y-Yo = Vot + 0.5at^2
43.9 = 0 + 0.5(9.8)t^2 = 4.9t^2 --> t1 = 2.99s

Since they both reached the water at the same time, t1 = 2.99s = t2.
Since stone 2 is thrown 1.00 s after stone 1, I decided to do: 2.99 + 1.00 = 3.99s to use for t in Y-Yo = Vot + 0.5at^2 for stone 2

So I got:

43.99 = Vo(3.99) + 0.5(-9.8)(3.99)^2 --> Vo = 30.5535 m/s

For some reason, I believe my method to be wrong because doesn't the t in Y-Yo = Vot + 0.5at^2 mean the total time it took to reach the water? And since they both reached the water at the same time, how can the t I used the second time be 3.99s? If I used 2.99s for t to get the Vo for stone 2, wouldn't that give me a Vo of 0? I am just confused as how to incorporate that 1.00s time lag and such :(

Also, because I've had much difficulty with these problems, can anyone give me some tips to solving these problems? Thanks so much!

you are confused by the time the second rock leaves. What you put says that the first rock reaches the water and then one second later you throw the second rock. Instead the first rock gets dropped, and while it is in the air, the second rock is thrown.

Here is the detailed solution:

For the first rock the equation is:

y = y0 - v0*t -(1/2)*a*t^2

y = 0m
y0 = 43.9m
v0 = 0m/s
t= ?
a = g

0 = 43.9 - (1/2)*g*t^2

Therefore the time it takes to get to the bottom is:

t = 2.99s

If the second rock left 1 second after, but both hit the water at the same time, the second rock only has 1.99s to reach the water

Therefore the equation for the second rock is

y = y0 - v0*t -(1/2)*a*t^2

y = 0m
y0 = 43.9m
v0 = ?
t= 1.99s
a = g

0 = 43.9 - v0*(1.99) - (1/2)*g*(1.99)^2

you can now solve for v0 which is:

v0 = 12.31 m/s

Last edited:
Missrikku: Since DDuardo gave the full answer (very nicely), you might have missed seeing exactly where you made your mistake:

The first stone takes 2.99 seconds to hit the water. You throw the second stone 1 second after the first and it hits the water at the same time as the first. That means the second stone falls for one second LESS than the first, not more! The "t" for the second stone is 2.99-1= 1.99 seconds NOT 2.99+1.

## 1. What is the concept of "Stone dropping - yet another free fall acceleration problem"?

The concept of "Stone dropping - yet another free fall acceleration problem" is a physics problem that involves calculating the acceleration of a falling object due to gravity. It is similar to other free fall acceleration problems, but may involve different variables or conditions.

## 2. How is this problem typically solved?

This problem is typically solved using the equations of motion, specifically the equation for free fall acceleration: a = g = 9.8 m/s². Other variables such as initial velocity, time, and displacement may also be used to solve the problem.

## 3. What factors affect the acceleration of a falling object in this problem?

The acceleration of a falling object in this problem is affected by two main factors: the force of gravity and any air resistance. The force of gravity is constant at 9.8 m/s², but air resistance can vary depending on the size and shape of the object.

## 4. Can this problem be applied to real-life scenarios?

Yes, this problem can be applied to real-life scenarios such as calculating the acceleration of a falling object, like a stone or a ball, from a certain height. It can also be used to predict the motion of objects in free fall, such as parachutes or skydivers.

## 5. Are there any limitations to this problem?

One limitation of this problem is that it assumes a vacuum environment, where there is no air resistance. In reality, air resistance can significantly affect the acceleration of a falling object. Additionally, this problem only considers the effects of gravity and does not take into account other forces that may be acting on the object.

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