ball thrown up with v=10m/s from a building of height 100. find max height and time taken to reach this max.

i said $\displaystyle a=-g $

integrate to get $\displaystyle v=-gt+C$ but initially v=10 so $\displaystyle v=-gt+10$

at max height v=0 so t=10/g

now i integrate again to get $\displaystyle s=-(gt^2)/2+10+K $but initially s=100 so K=100

so at t=10/g, i get s= 5.1

this answer is correct but they use

$\displaystyle v_{0}^2-2g \Delta h=v$

can anyone show me how they got this equation?

For part b) they want the time taken to go from h=50 to h=0.

for this i used the formula above for s and solved for t wneh h-50 and 0 to get the correct answer.

in the solutions they use

$\displaystyle v_{0}^2-2g \Delta h=v$

and if v2 is velocity at H=0 and v1 is velocity at H=50 then

v2=v1-gt

is the above obvious?im thinking,wrongly, if i start motion at v1 then f=ma gives me a=g so v=gt+K but initially v=v1 thus v=gt+v1 ??

Can i get v2=v1-gt starting with f=ma?