A particle is thrown up inside a stationary lift of sufficient height. The time of flight is T. Now it is thrown again with same initial speed v0 with respect to lift. At the time of second throw, lift is moving up with speed v0 and uniform acceleration g upward (the acceleration due to gravity) The new time of flight is

Question

A particle is thrown up inside a stationary lift of sufficient height. The time of flight is T. Now it is thrown again with same initial speed v0 with respect to lift. At the time of second throw, lift is moving up with speed v0 and uniform acceleration g upward (the acceleration due to gravity) The new time of flight is (A) √2 T (B) (T/2) (C) T (D) 2T

Tardigrade Admin · Accepted Answer

With respect to lift initial speed =v0 Acceleration = -2 g Displacement = 0 ∴ S=ut+(1/2)at2 0=v0T'+(1/2)× 2g× T'2 ∴ T'=(v0/g)=(1/2)×(2v0/g)=(1/2) T

$2 T$

$\frac{T}{2}$

$T$

$2 T$

With respect to lift initial speed $= v_{0}$
Acceleration $= - 2 g$
Displacement $= 0$
$∴ S = u t + \frac{1}{2} a t^{2}$
$0 = v_{0} T^{'} + \frac{1}{2} \times 2 g \times T^{' 2}$
$∴ T^{'} = \frac{v _{0}}{g} = \frac{1}{2} \times \frac{2 v _{0}}{g} = \frac{1}{2} T$

2​T

2T​

T

2T

With respect to lift initial speed =v0​ Acceleration =−2g Displacement =0 ∴S=ut+21​at2 0=v0​T′+21​×2g×T′2 ∴T′=gv0​​=21​×g2v0​​=21​T

$2 T$

$\frac{T}{2}$

$T$

$2 T$

With respect to lift initial speed $= v_{0}$
Acceleration $= - 2 g$
Displacement $= 0$
$∴ S = u t + \frac{1}{2} a t^{2}$
$0 = v_{0} T^{'} + \frac{1}{2} \times 2 g \times T^{' 2}$
$∴ T^{'} = \frac{v _{0}}{g} = \frac{1}{2} \times \frac{2 v _{0}}{g} = \frac{1}{2} T$