Question

An object falls freely from 180 m height

a) Find the velocity with which it strikes the ground. 1

b) Find the time taken to reach the ground. 1

c) Find the displacement in 3rd second of falling 1

d) Write the displacement-time relation​

Answer 1

Explanation:

h=180m

u=0

g =9.8

then according v2=u2+2gh

v2=02+2×9.8×180

=3528

v2= 3528

v = 59.39.....

Answer 2

$\huge\bigstar \: \huge\mathrm { \underline{ \purple{given} }} \: \bigstar \: </p><p> \:$

• u = 0m/sec ( free fall )
• v = ?
• s = 180m
• g = 10m/sec²

$\\ \\ \huge\bigstar \: \huge\mathrm { \underline{ \purple{solution} }} \: \bigstar \: \\ \\ </p><p>$

${ \bf{ \red{a)}} }\rm{velocity \: at \: which \: it \: strikes \: the \: ground.}$

➥ By 3rd Kinematical Equation,

$\boxed{ \pink{ {v}^{2} = {u}^{2} + 2as}}$

➣ Putting values we get,

$\sf{ {v}^{2} = {0}^{2} + 2(10)(180)} \\ \\ \sf{ {v}^{2} = 3600 } \\ \\ \sf{ v = \sqrt{3600} } \\ \\ \large\sf \boxed{ \sf{ v =60 \frac{m}{sec} }}$

$\therefore \sf{the \: velocity \: at \: which \: the \: ball \: strikes} \\ \sf{the \: ground \: is} \: \blue{60 \frac{m}{sec} } \: .$

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$\bf{ \red{b )}} \rm{time \: taken \: to \: reach \: the \: ground.}$

➥ By 1st Kinematical Equation,

$\boxed{ \pink{v = u + gt}}$

➣ Putting values we get,

$\sf{60 = 0 + (10)(t)} \\ \\ \sf{60 = 10t} \\ \\ \large\boxed{ \sf {t = 6sec}}$

$\therefore \sf{time \: taken \: to \: reach \: the \: ground \: is \: } \\ \: \: \: \: \sf{ \blue{6sec}} \: .$

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$\large \bf{ \red{c)}} \rm{displacement \: in \: 3rd \: sec.}$

Displacement in 3rd second

ㅤㅤ ㅤㅤ ㅤ= displacemnt in 3sec - displacement in 2sec

➥ By 2nd Kinematical equation,

$\boxed{ \pink{s = ut + \frac{1}{2} a {t}^{2} }}$

➽ We will find displacement in 3sec.

$\sf{ s \: _{(3sec)} = (0)(3) + \frac{1}{2} (10)( {3)}^{2} } \\ \\ \sf{ s \: _{(3sec)} \: = \frac{1}{2}(10)(9) } \\ \\ \sf{ s \: _{(3sec)} \: = 45m}$

➽ Now, we will find displacement in 2sec.

$\sf{ s \: _{(2sec)} \: = (0)(2) + \frac{1}{2}(10)( {2)}^{2} } \\ \\ \sf{ s \: _{(2sec)} \: = \frac{1}{2} (10)(4)} \\ \\ \sf{ s \: _{(2sec)} \: = 20m}$

Displacement in 3rd sec = 45-20 = 25m

$\therefore \rm{displacement \: in \: 3rd \: sec \: is \:} \blue{25m}.$

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$\large \bf{ \red{d)}} \rm{displacement \: time \: relation.}$

$\sf{s = ut + \frac{1}{2} a {t}^{2} } \\ \\ \sf{s = (0)t + \frac{1}{2} (10)( {t)}^{2} } \\ \\ \large\boxed{ \sf{s = 5 {t}^{2} }}$

$\therefore \rm{displacement \: time \: relation \: is \: } \\ \: \: \: \: \: \sf{ \blue{s = 5 {t}^{2} }}.$