Verified Solution • 2025–26 Session moderate

QJEEM22Phdbb0: JEE Main 2022 Physics - Class 11 JEEM22Phdbb0 Solution: From the top of a tower, a ball is thrown vertically upward…

From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in ____________ s.

Step-by-Step Verified Solution

Based on the situation

h=−ut1+12gt12h = - u{t_1} + {1 \over 2}gt_1^2 →\to throwing up ....... (i)

h=ut2+12gt22h = u{t_2} + {1 \over 2}gt_2^2 →\to throwing up ....... (ii)

h=12gt2h = {1 \over 2}g{t^2} →\to dropping .......... (iii)

and 0=u(t1−t2)−12g(t1−t2)20 = u({t_1} - {t_2}) - {1 \over 2}g{({t_1} - {t_2})^2} ....... (iv)

solving above equations

t=t1t2t = \sqrt {{t_1}{t_2}}

⇒t=6×1.5=3 s\Rightarrow t = \sqrt {6 \times 1.5} = 3\,s

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