Question on an object falling 10 m

[pureouchies4717]

vi= 20m/s
t=?
a=9.8m/s^2
vf=?
d=-10

here is the question:

A rock is tossed straight up with a velocity of + 20m/s When it returns, it falls into a hole 10m deep.

What is the rock's velocity as it hits the bottom of the hole?

i did the vf^2=vi^2+2ad equation

and got: 24.41, but its wrong

 

[mrjeffy321]

In this case, the distance d would be the entire distance the rock travels while it is accelerating (due to gravity), so it would be the distance up, down, and then down some more.

You can also simplify this question.
If the rock is thrown up 20 m/s, it eventually reaches a maximum height and falls back down the same distance. When the rock reaches the height it originally was launches from (in this case 0), it is going the same speed as when it left (but in the opposite direction).
By knowing this, you can just start the rock at zero height and traveling -20 m/s, then use the 10 meters it falls down as the distance to find its final velocity.

 

[pureouchies4717]

thanks for your help. i got this as my answer: 31.56m/s

is that right?

 

[andrewchang]

i got your original answer, 24.41m/s ...

 

[G01]

I also got your original answer. And another one? Must have made a mistake...

 

[sporkstorms]

If you're uncomfortable using the conclusion mrjeffy made, break the problem up into finer steps: 1) Going up, then 2) coming down.

Throughout each of these steps, acceleration is constant and remains a=g~=-9.8m/s^2

1a) Find the time (call it t_1) it takes for the rock to go as high as it will go:
When the rock starts it has v_i1=20m/s. When it stops at the top and turns to come back down, it has a v_f1=0m/s.

$$v_{f,1} = v_{i,1} + at_1$$

1b) using this t_1, v_i1, v_f1 and a: find the distance the rock traveled, call it d_1.

$$d_1 = v_{i,1} t_1 + \frac{1}{2}at_1^2$$

2a) Now the rock is still at the top of the path. We know it's d_1 m above the ground. It now has to fall d_2 = d_1 + 10 m. It starts out at v_i2 = 0m/s (suspended in the air momentarily, before it turns and heads down).

Use d_2, v_i2, and a to find t_2, the time it takes the rock to fall a distance of d_2 (from the top, down to the bottom of the hole).

$$d_2 = v_{i,2} t_2 + \frac{1}{2}at_2^2$$

2b) Now that we know the time it takes we can find v_f2, the velocity of the rock at the bottom of the hole.

$$v_{f,2} = v_{i,2} + at_2$$

 

[G01]

I figured out my stupid mistake. I also got your original answer nick.

 

[pureouchies4717]

If you're uncomfortable using the conclusion mrjeffy made, break the problem up into finer steps: 1) Going up, then 2) coming down.

Throughout each of these steps, acceleration is constant and remains a=g~=-9.8m/s^2

1a) Find the time (call it t_1) it takes for the rock to go as high as it will go:
When the rock starts it has v_i1=20m/s. When it stops at the top and turns to come back down, it has a v_f1=0m/s.

$$v_{f,1} = v_{i,1} + at_1$$

1b) using this t_1, v_i1, v_f1 and a: find the distance the rock traveled, call it d_1.

$$d_1 = v_{i,1} t_1 + \frac{1}{2}at_1^2$$

2a) Now the rock is still at the top of the path. We know it's d_1 m above the ground. It now has to fall d_2 = d_1 + 10 m. It starts out at v_i2 = 0m/s (suspended in the air momentarily, before it turns and heads down).

Use d_2, v_i2, and a to find t_2, the time it takes the rock to fall a distance of d_2 (from the top, down to the bottom of the hole).

$$d_2 = v_{i,2} t_2 + \frac{1}{2}at_2^2$$

2b) Now that we know the time it takes we can find v_f2, the velocity of the rock at the bottom of the hole.

$$v_{f,2} = v_{i,2} + at_2$$

thanks so much, and sorry to waste your time

 

[pureouchies4717]

correct answer: -24.4 m/s

but now its asking

How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

to this question, i got 1.77s, but this is also wrong. ugh why does physics have to be so hard

 

[sporkstorms]

How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

Well, if you follow the steps listed above, all you have to do is add t_1 and t_2.

One nice thing about physics is that, often, by taking the steps to find the answer to one question, you find the answers to 30 more questions along the way.

 

[pureouchies4717]

would the time be equal to .4183?

please let me know, this is the last time i could give an answer, and if its wrong i get no credit at all despite my hard work

 

[sporkstorms]

would the time be equal to .4183?

please let me know, this is the last time i could give an answer, and if its wrong i get no credit at all despite my hard work

Follow the steps and solve the problem fully.
There are also several ways to check your answer, if you understand what's going on with each of the steps.

If you don't understand, ask about the parts that don't make sense.

edit: but no, that's not the right answer. If you continue to get the wrong answer, write out the steps you took and why you took them, and I'll be glad to show you where it went wrong.

 

[pureouchies4717]

are you guys allowed to confirm a right answer?

 

[moresko]

Solution

Hi,
You should try breaking the questions into 2 parts (going up and coming down), the time going up is 2.04s and the time coming down is 2.49s. Add this numbers and get your answer.