Question

A stone projected vertically up from the top of a cliff reaches the foot of the cliff in $8s$. If it is projected vertically downwards with the same speed, it reaches the foot of the cliff in $2s$. Then its time of free fall from the cliff is

• A. $16s$
• B. $8s$
• C. $2s$
• D. $4s$

Answer 1

Answer: (D)

$4s$

$A$ to $B$ $\to$ (with u)
$\Rightarrow H=u\left(8\right)-\frac{1}{2}g(8{)}^2=8u-32g-(1)$
$A$ to $B$ (with u downwords)
$\Rightarrow H=u\left(2\right)-\frac{1}{2}g(2{)}^2=2u-2g-(2)$

Free fall, $e{q}^n\left(1\right)-e{q}^n\left(2\right)$:
$t=\sqrt{\frac{2H}{g}}\Rightarrow 6u=30g$
$u=5g$

So, $H=10g-2g=8g$(from 2)
Hence, $t=\sqrt{\frac{2\times 8g}{g}}=4sec$