What is the Solution for Finding Ball's Speed at Impact and Maximum Height?

[ScullyX51]

Homework Statement

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.

Homework Equations

E_total= PE_i+KE_i=PE_f+KE_f
Vf²-Vi²=2AD
KE=mv²
PE=mgy
h_max= v²/ 2g

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.

 

[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.

 

[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.

 

[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:
v_f=final velocity
v_i=initial velocity
a=acceleration due to gravity
\Delta y=height above ground

 

[ScullyX51]

Vf²-vi²=2ad
vf²=2ad
(I set vi=0 since it starting at hmax)
Vf= sqrt( 2*9.8*25.1)=22.2 m/s

 

[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.
Possibilities of error in your answer could be due to:
1) the question takes air resistance into account
2) gravity is something other than 9.8ms^-2
3) Misread/misinterpreted the question.