Verified Solution • 2025–26 Session moderate

QJEEM20Ph7c33: JEE Main 2020 Physics - Class 11 JEEM20Ph7c33 Solution: A helicopter rises from rest on the ground vertically upwards…

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]

Multiple Choice Options

  • A t = 3.4 (hg)\sqrt {\left( {{h \over g}} \right)}
  • B t = 1.8 (hg)\sqrt {\left( {{h \over g}} \right)}
  • C t = 2h3g\sqrt {{{2h} \over {3g}}}
  • D t = 23(hg){2 \over 3}\sqrt {\left( {{h \over g}} \right)}

Step-by-Step Verified Solution

JEE Main 2020 (Online) 5th September Morning Slot Physics - Motion in a Straight Line Question 77 English Explanation
Velocity of helicopter at height h,

VB2=02+2ghV_B^2 = {0^2} + 2gh

VB=2gh{V_B} = \sqrt {2gh}

−h=(VB)t−12gt2- h = ({V_B})t - {1 \over 2}g{t^2}

⇒\Rightarrow −h=2ght−12gt2- h = \sqrt {2ght} - {1 \over 2}g{t^2}

⇒\Rightarrow gt2−22ght−2h=0g{t^2} - 2\sqrt {2ght} - 2h = 0

⇒\Rightarrow t=2ght±8gh+8gh2g=22gh±16gh2g=2gh+2ghgt = {{\sqrt {2ght} \pm \sqrt {8gh + 8gh} } \over {2g}} = {{2\sqrt {2gh} \pm \sqrt {16gh} } \over {2g}} = {{\sqrt {2gh} + 2\sqrt {gh} } \over g}

⇒\Rightarrow t=2hg+2hg=hg(2+2)=3.4hgt = \sqrt {{{2h} \over g}} + 2\sqrt {{h \over g}} = \sqrt {{h \over g}} \left( {\sqrt 2 + 2} \right) = 3.4\sqrt {{h \over g}}

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