Question

A stone dropped from the top of a building has a velocity of 20m/s on reaching ground level. Find the time for which the stone is falling freely.

Answer 1

$\underline{ \underline{ \huge{ \pmb{ \sf {Given :}}}}}$

• Initial velocity (u) = 0 m/s

• Final velocity (v) = 20 m/s

$\underline{ \underline{ \huge{ \pmb{ \sf {To \: calculate :}}}}}$

• Time (t)

$\underline{ \underline{ \huge{ \pmb{ \sf {Calculation :}}}}}$

By using the first equation of motion for freély falling bodies,

• $\underline{ \boxed{ \large{ \pmb{ \sf { v = u + gt}}}}}$

Where,

◍ v denotes final velocity.

◍ u denotes initial velocity.

◍ g denotes acceleration due to gravity.

◍ t denotes time.

Here, $\boxed{\sf{g = 10 \: {m/s}^{2} }}$

Now, substituting values :

$\longrightarrow \sf { v = u + gt}$

$\longrightarrow \sf { 20 = 0 + 10t}$

$\longrightarrow \sf { 20 -0 = 10t}$

$\longrightarrow \sf { 20= 10t}$

$\longrightarrow \sf { \dfrac{20}{10} = t}$

$\longrightarrow \underline{\boxed{\sf { 2\: s = t}}}$

• Henceforth, for 2 seconds the stone was falling freely.

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More Formulae:

First equation of motion :

• v = u + at

Second equation of motion :

• s = ut + ½at²

Third equation of motion :

• v² - u² = 2as

Where,

◍ v denotes final velocity. [ in m/s ]

◍ u denotes initial velocity. [ in m/s ]

◍ a or g denotes acceleration or acceleration due to gravity. [ in m/s² ]

◍ t denotes time. [ in seconds ]

◍ s denotes distance. [ in m(metres)]

Answer 2

Final velocity = 20 m/s

Initial velocity = 0 m/s

Acceleration = g = 10 m/s²

We know,

v = u + gt (1st equation)

⇒ t = (v - u)/g = v/g

⇒ t = (20 m/s)/(10 m/s²)

⇒ t = 2 s

Hence, it took 2 seconds to descent.