Question 66: Motion in 1D

Question

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]

Accepted Answer

Velocity of helicopter at height h,
$$V_B^2 = {0^2} + 2gh$$
$${V_B} = \sqrt {2gh} $$
$$ - h = ({V_B})t - {1 \over 2}g{t^2}$$
$$ \Rightarrow $$ $$ - h = \sqrt {2ght}  - {1 \over 2}g{t^2}$$
$$ \Rightarrow $$ $$g{t^2} - 2\sqrt {2ght}  - 2h = 0$$
$$ \Rightarrow $$ $$t = {{\sqrt {2ght}  \pm \sqrt {8gh + 8gh} } \over {2g}} = {{2\sqrt {2gh}  \pm \sqrt {16gh} } \over {2g}} = {{\sqrt {2gh}  + 2\sqrt {gh} } \over g}$$
$$ \Rightarrow $$ $$t = \sqrt {{{2h} \over g}}  + 2\sqrt {{h \over g}}  = \sqrt {{h \over g}} \left( {\sqrt 2  + 2} \right) = 3.4\sqrt {{h \over g}} $$

Answer

t = 3.4$$\sqrt {\left( {{h \over g}} \right)} $$

Answer

t = 1.8$$\sqrt {\left( {{h \over g}} \right)} $$

Answer

t = $$\sqrt {{{2h} \over {3g}}} $$

Answer

t = $${2 \over 3}\sqrt {\left( {{h \over g}} \right)} $$

Motion in 1D

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]

Choose the correct answer: