
⇒For upward motion of helicopter
v2=u2+2as
v2=0+2gh
v=2gh
⇒ Now particle will start moving under gravity.
s=ut+21at2
−h=2ght−21gt2
then t=2×2g2gh±2gh+4×2g×h
t=g2gh(1+2)
t=g2h(1+2)
JEE Main 2020 — Physics Mechanics
A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height h. The time taken by the packet to reach the ground is close to [g is the acceleration due to gravity]:
Held on 5 Sept 2020 · Verified 6 Jul 2026.
t=32(gh)
t=1.8(gh)
t=3.4(gh)
t=3g2h
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