If a particle takes t second less and acquires a velocity of v m/s more in falling through the same distance on two planets where the accelerations due to gravity are 2 g and 8 g respectively then

Question

If a particle takes t second less and acquires a velocity of v m/s more in falling through the same distance on two planets where the accelerations due to gravity are 2 g and 8 g respectively then (A) v = 4 gt (B) v = 5 gt (C) v = 2 gt (D) v = 16 gr

Tardigrade Admin · Accepted Answer

(v/t)=(v1-v2/t2-t1) =(√2 a1 s-√2 a2 s/√(2 s)a2-√(2 s)a1 ⇒ v=(√/a1 a2)) t ⇒ v=√(2 g)(8 g) t ⇒ v=√16 g2 t ⇒ v=4 g t

Q. If a particle takes $t$ second less and acquires a velocity of $v m / s$ more in falling through the same distance on two planets where the accelerations due to gravity are $2 g$ and $8 g$ respectively then

v = 4 gt

v = 5 gt

v = 2 gt

v = 16 gr

$\frac{v}{t} = \frac{v _{1} - v _{2}}{t _{2} - t _{1}}$
$= \frac{2 a _{1} s - 2 a _{2} s}{\frac{2 s}{a _{2}} - \frac{2 s}{a _{1}}}$
$\Rightarrow v = (a_{1} a_{2}) t$
$\Rightarrow v = (2 g) (8 g) t$
$\Rightarrow v = 16 g^{2} t$
$\Rightarrow v = 4 g t$

Q. If a particle takes t second less and acquires a velocity of vm/s more in falling through the same distance on two planets where the accelerations due to gravity are 2g and 8g respectively then