Two balls of different masses ma mb are dropped from two different heights a and b. The ratio of the time taken by the two to cover these distances are:

Question

Two balls of different masses ma mb are dropped from two different heights a and b. The ratio of the time taken by the two to cover these distances are: (A) 1 (B) √a / b (C) b: a (D) a: b

Tardigrade Admin · Accepted Answer

According to the relation h = ut +(1/2) gt 2 ] h =(1/2) gt 2 ( u =0) h propto t2 ∴ t propto √ h Hence, (ta /tb)=√(a/3)

Q. Two balls of different masses $m_{a}$ & $m_{b}$ are dropped from two different heights $a$ and $b$ . The ratio of the time taken by the two to cover these distances are:

$1$

$a / b$

$b : a$

$a : b$

According to the relation $h = u t + \frac{1}{2} g t^{2}$ ]
$h = \frac{1}{2} g t^{2} (u = 0)$
$h \propto t^{2}$
$∴ t \propto h$
Hence,

$\frac{t _{a}}{t _{b}} = \frac{a}{3}$

Q. Two balls of different masses ma​ & mb​ are dropped from two different heights a and b. The ratio of the time taken by the two to cover these distances are:

1

a/b​

b:a

a:b

According to the relation h=ut+21​gt2 ] h=21​gt2(u=0) h∝t2 ∴t∝h​ Hence, tb​ta​​=3a​​

Q. Two balls of different masses $m_{a}$ & $m_{b}$ are dropped from two different heights $a$ and $b$ . The ratio of the time taken by the two to cover these distances are:

$1$

$a / b$

$b : a$

$a : b$

According to the relation $h = u t + \frac{1}{2} g t^{2}$ ]
$h = \frac{1}{2} g t^{2} (u = 0)$
$h \propto t^{2}$
$∴ t \propto h$
Hence,

$\frac{t _{a}}{t _{b}} = \frac{a}{3}$