Question

KINEMATICS - Motion in a Straight line
Iwo bodies A of mass 1 kg and B of mass 3
are dropped from heights of 16 m and 25 m respec
ively. Calculate the ratio of the time taken by them
o reach the ground.
PSEB​

Answer 1

${ \red{ \bf{ \underline{ \purple{ \underline{Given \: data}}}}}:-}$

→ Mass of body A = 1 kg

→ Mass of body B = 3 kg

→ Body A dropped from height (h1) = 16 m

→ Body B dropped from height (h2)= 25 m

${ \red{ \bf{ \underline{ \purple{ \underline{Solution}}}}}:-}</p><p>$

Let, initial velocity (u) of both bodies be zero ( 0 ) .

Here, a is acceleration of the particle, t is time taken by body, s is displacement of body and u is intial velocity of body.

Now, we use kinemetical equations

${ \bf{ \dashrightarrow{s = ut + \frac{1}{2} a {t}^{2} }}}$

→ Let, h is displacement of body

Hence, s = h & take h = h1

${ \bf{ \dashrightarrow{h1 = ut + \frac{1}{2} a {t}^{2} }}}$

according to assumption u = 0, hence

${ \bf{ \dashrightarrow{h1 = (0)t + \frac{1}{2} a {t}^{2} }}}$

${ \bf{ \dashrightarrow{h1 = \frac{1}{2} a {t}^{2} }}}$

Here, we know acceleration of body is equal to acceleration due to gravity.

Hence, a = g and we know g is constant

g = 9.8 m/s².

${ \bf{ \dashrightarrow{h1 = \frac{1}{2} g {t}^{2} }}}$

Here g and ½ constant hence,

${ \bf{ \dashrightarrow{h1 = {(t_{1} )}^{2} }}}$

similarly,

${ \bf{ \dashrightarrow{h2= {(t_{2})}^{2} }}}$

Now, for ratio of time taken by bodies

${ \bf{ \dashrightarrow{ \frac{ h1}{h2} = \frac{ {(t_{1})}^{2} }{ ({t_{2})}^{2} } }}}$

${ \bf{ \dashrightarrow{ ({\frac{ t_{1} }{ t_{2}})}^{2} = \frac{h1}{h2} }}}$

${ \bf{ \dashrightarrow{ \frac{ t_{1} }{ t_{2}} = \sqrt{\frac{h1}{h2} } }}}$

${ \bf{ \dashrightarrow{ \frac{ t_{1} }{ t_{2}} = \sqrt{\frac{16}{25} } }}}$

${ \bf{ \dashrightarrow{ \frac{ t_{1} }{ t_{2}} = \frac{4}{5} }}}$

Hence, the ratio of the time taken by bodies to reach the ground is 4/5.