
The maximum height of a projectile thrown vertically upward is given by, h=2gu2.
⇒u=2gh
Using equation of motion with constant acceleration,
s=ut+21at2
⇒3h=2ght−21gt2
⇒2gt2−2ght+3h=0
Solving the above quadratic equation we will get the times t1 and t2. ⇒t1t2=2gh−2gh−4×2g×3h2gh+2gh−4×2g×3h=2gh−34gh2gh+34gh=3−23+2
Thus, the ratio t2t1=3+23−2