Question

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

• A. v = 4gt
• B. v = 5gt
• C. v = 2gt
• D. v = 16gt

Answer 1

The correct option is A v = 4gt

In first planet

$a=2g$

Time taken $=T$

Distance $=\frac{1}{2}2g{T}^2=g{T}^2$

Velocity achieved $=aT=2gT$

In second planet

$a=8g$

Time taken $=T-t$

Distance $=\frac{1}{2}8g(T-t{)}^2=4g(T-t{)}^2$

Velocity achieved $=8g(T-t)$

$8g(T-t)=2gT+V$

$⟹6gT-8gt=V$ ........(1)

Since the distance is same,

$g{T}^2=4g(T-t{)}^2$

$⟹T^2=4(T-t{)}^2$

$⟹T=2(T-t)$

$⟹T=2T-2t$

$⟹T=2t$

Putting the value of $T$ in equation (1), we get

$6g\left(2t\right)-8gt=V$

$⟹V=4gt$