corpul va urca pana cand v=0
adica vo-gt=0
20-10t=0
t=2
timp in care va fi la inaltimea
vm*t=vo*t/2=20*2/2=20m deci la 20 m ajunge in 2secunde
altfel
vot-gt²/2=20
20t-5t²=20
4t-t²=4
t²-4t+4=0
t1=t2=2sec
solutie unica dubla e la maxim (varful parabolei ce reprezinmta functiade gradul 2 a distantei)
dar la
la 10m??
vot-gt²/2=10
20t-5t²=10
4t-t²=2
t²-4t+2=0
t1,2=(4+-√(16-8))/2
t1,2=(4+-2√2)/2
t1,2=2+-√2
t1=2-√2 sec≈0,7secunde LA URCARE
t2=2+√2 secunde≈3,4 secunde LA COBOARARE )
