When two particles each of mass m are dropped from height h and 2h respectively, then the ratio of their times to reach the ground is

Question

When two particles each of mass m are dropped from height h and 2h respectively, then the ratio of their times to reach the ground is (A)  1:√2  (B)  √2:1  (C)  1:2  (D)  2:1

Tardigrade Admin · Accepted Answer

From relation h=u t +(1/2) g t2 (with u=0), we have h=(1/2) g t2 ⇒ t=√((2 h/g)) propto √h ∴ (t1/t2)=√(h1/h2) √((h/2 h))=(1/√2)

Q. When two particles each of mass $m$ are dropped from height h and $2 h$ respectively, then the ratio of their times to reach the ground is

$1 : 2$

$2 : 1$

$1 : 2$

$2 : 1$

From relation
$h = u t + \frac{1}{2} g t^{2}$ (with $u = 0)$ ,
we have $h = \frac{1}{2} g t^{2}$
$\Rightarrow t = (\frac{2 h}{g}) \propto h$
$∴ \frac{t _{1}}{t _{2}} = \frac{h _{1}}{h _{2}} (\frac{h}{2 h}) = \frac{1}{2}$

Q. When two particles each of mass m are dropped from height h and 2h respectively, then the ratio of their times to reach the ground is

1:2​

2​:1

1:2

2:1

From relation h=ut+21​gt2 (with u=0), we have h=21​gt2 ⇒t=(g2h​)​∝h​ ∴t2​t1​​=h2​h1​​​(2hh​)​=2​1​

Q. When two particles each of mass $m$ are dropped from height h and $2 h$ respectively, then the ratio of their times to reach the ground is

$1 : 2$

$2 : 1$

$1 : 2$

$2 : 1$

From relation
$h = u t + \frac{1}{2} g t^{2}$ (with $u = 0)$ ,
we have $h = \frac{1}{2} g t^{2}$
$\Rightarrow t = (\frac{2 h}{g}) \propto h$
$∴ \frac{t _{1}}{t _{2}} = \frac{h _{1}}{h _{2}} (\frac{h}{2 h}) = \frac{1}{2}$