A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet…

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]:
  1. t=23hg
  2. t=1.8hg
  3. t=3.4hg
  4. t=2h3g

Solution

For upward motion of helicopter

v2=u2+2as

v2=0+2gh

v=2gh

 Now particle will start moving under gravity.

s=ut+12at2

-h=2ght-12gt2

then t=2gh±2gh+4×g2×h2×g2

t=2ghg(1+2)

t=2hg(1+2)
 

Asked in: JEE Main 2020 (05 Sep Shift 1)

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