- Home
- Standard 11
- Physics
2.Motion in Straight Line
hard
If a particle takes $t$ second less and acquires a velocity of $v \ ms^{^{-1}}$ more in falling through the same distance (starting from rest) on two planets where the accelerations due to gravity are $2 \,\, g$ and $8 \,\,g$ respectively then $v=$
A$v = 2gt$
B$v = 4gt$
C$v = 5 gt$
D$v = 16 gt$
Solution
$\frac{v}{t}=\frac{v_{1}-v_{2}}{t_{2}-t_{1}}=\frac{\sqrt{2 a_{1} s}-\sqrt{2 a_{2} s}}{\sqrt{\frac{2 s}{a_{2}}}-\sqrt{\frac{2 s}{a_{1}}}}$
$\Rightarrow$ $v=(\sqrt{a_{1} a_{2}}) t$
$\Rightarrow v=\sqrt{(2 g)}(8 g) t$
$\Rightarrow$ $=\sqrt{16 g^{2} t}$
$\Rightarrow v=4 g t$
$\Rightarrow$ $v=(\sqrt{a_{1} a_{2}}) t$
$\Rightarrow v=\sqrt{(2 g)}(8 g) t$
$\Rightarrow$ $=\sqrt{16 g^{2} t}$
$\Rightarrow v=4 g t$
Standard 11
Physics
Similar Questions
hard
