# Find the Speed at Impact

1. Nov 18, 2008

### ScullyX51

1. The problem statement, all variables and given/known data
A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he releases it. Use energy to find:
1)The ball's maximum height above the ground.
2)The ball's speed as it passes the window on its way down.
3)The speed of impact on the ground.

2. Relevant equations
Etotal= PEi+KEi=PEf+KEf
KE=mv2
PE=mgy
hmax= v2/ 2g

3. The attempt at a solution
I solved for the first part and got it right by using the formula for hmax and got 5.1 ,and then added 20 m and got 25.1 meters which is correct.
For the second part I used vf2-vi2=2ad
and I got 10 m which is also correct.
I am having trouble finding the third one; I tried applying vf2 -vi2=2ad using 20 m and 25.1 m and both of them didn't work.
any suggestions.

2. Nov 18, 2008

### rl.bhat

When you through a body upwards, the body returns back to the point of projection with the velocity which is equal to the velocity of projection.

3. Nov 18, 2008

### ScullyX51

Then if it is the same as the point of projection it would be 10 m/s, and mastering physics is saying that is wrong.

4. Nov 18, 2008

### Mentallic

If you show us how you went about calculating the last question, we can direct you to the problem.

Using $$v_{f}^{2}-v_{i}^{2}=2a\Delta y$$ where:
vf=final velocity
vi=initial velocity
a=acceleration due to gravity
$$\Delta y$$=height above ground

5. Nov 18, 2008

### ScullyX51

(I set vi=0 since it starting at hmax)
Vf= sqrt( 2*9.8*25.1)=22.2 m/s

6. Nov 18, 2008

### Mentallic

Yes. I can, with almost certainty, tell you that this IS the correct answer and if the printed answer you are trying to achieve is well off this one you are getting, then you are right and the answer in the book is wrong.