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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 on two planets, where the accelerations due to gravity are $2\,g$ and $8\,g$ respectively, then
A$v=4 g t$
B$v=5 g t$
C$v=2 g t$
D$v=16 g t$
Solution
(a)
Time $=\sqrt{\frac{2 h}{g}}$ and velocity $=\sqrt{2 g h}$
Now, $t=\sqrt{\frac{2 h}{2 g}}-\sqrt{\frac{2 h}{8 g}}$ and $v=\sqrt{16 g h}-\sqrt{4 g h}$
Dividing these two equations, we get
$\frac{t}{v}=\frac{\sqrt{1 / 2\,g}-\sqrt{1 / 8\,g}}{\sqrt{8\,g}-\sqrt{2\,g}}$
$=\frac{2-1}{8 g-4 g}=\frac{1}{4 g} \quad$ or $\quad v=4 g t$
Time $=\sqrt{\frac{2 h}{g}}$ and velocity $=\sqrt{2 g h}$
Now, $t=\sqrt{\frac{2 h}{2 g}}-\sqrt{\frac{2 h}{8 g}}$ and $v=\sqrt{16 g h}-\sqrt{4 g h}$
Dividing these two equations, we get
$\frac{t}{v}=\frac{\sqrt{1 / 2\,g}-\sqrt{1 / 8\,g}}{\sqrt{8\,g}-\sqrt{2\,g}}$
$=\frac{2-1}{8 g-4 g}=\frac{1}{4 g} \quad$ or $\quad v=4 g t$
Standard 11
Physics
