JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 8)
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]
[g is the acceleration due to gravity]
t = 3.4$$\sqrt {\left( {{h \over g}} \right)} $$
t = 1.8$$\sqrt {\left( {{h \over g}} \right)} $$
t = $$\sqrt {{{2h} \over {3g}}} $$
t = $${2 \over 3}\sqrt {\left( {{h \over g}} \right)} $$
Explanation
_5th_September_Morning_Slot_en_8_2.png)
Velocity of helicopter at height h,
$$V_B^2 = {0^2} + 2gh$$
$${V_B} = \sqrt {2gh} $$
$$ - h = ({V_B})t - {1 \over 2}g{t^2}$$
$$ \Rightarrow $$ $$ - h = \sqrt {2ght} - {1 \over 2}g{t^2}$$
$$ \Rightarrow $$ $$g{t^2} - 2\sqrt {2ght} - 2h = 0$$
$$ \Rightarrow $$ $$t = {{\sqrt {2ght} \pm \sqrt {8gh + 8gh} } \over {2g}} = {{2\sqrt {2gh} \pm \sqrt {16gh} } \over {2g}} = {{\sqrt {2gh} + 2\sqrt {gh} } \over g}$$
$$ \Rightarrow $$ $$t = \sqrt {{{2h} \over g}} + 2\sqrt {{h \over g}} = \sqrt {{h \over g}} \left( {\sqrt 2 + 2} \right) = 3.4\sqrt {{h \over g}} $$
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