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 a height h. The time taken by the packet to reach the ground is close to [ g is the acceleration due to gravity] :

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 a height h. The time taken by the packet to reach the ground is close to [ g is the acceleration due to gravity] : (A) t =√(2 h /3 g ) (B) t =1.8 √( h / g ) (C) t =3.4 √(( h / g )) (D) t =(2/3) √(( h / g ))

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/fecda85b61db34a5fc00a64f79da2fb0-.png" /> v2=u2+2 a s v2=2 g h+2 g h v=√4 gh ⇒ √4 gh =√2 gh + gt ⇒ t =√(4 h / g )-√(2 h / g ) ⇒ 3.4 √( h / g )

Q. 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 a height $h$ . The time taken by the packet to reach the ground is close to $[g$ is the acceleration due to gravity] :

$t = \frac{2 h}{3 g}$

$t = 1.8 \frac{h}{g}$

$t = 3.4 (\frac{h}{g})$

$t = \frac{2}{3} (\frac{h}{g})$

$v^{2} = u^{2} + 2 a s$
$v^{2} = 2 g h + 2 g h$
$v = 4 g h$
$\Rightarrow 4 g h = 2 g h + g t$
$\Rightarrow t = \frac{4 h}{g} - \frac{2 h}{g} < b r / >< b r / >\Rightarrow 3.4 \frac{h}{g}$

Q. 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 a height h. The time taken by the packet to reach the ground is close to [g is the acceleration due to gravity] :

t=3g2h​​

t=1.8gh​​

t=3.4(gh​)​

t=32​(gh​)​