Question

A stone is dropped from top of a tower of height 180 m above the ground if g=10 m/s² then time taken by the stone to reach on the ground is?​

Answer 1

⚘ Given That :

• ↣ Height of Tower (H) = 180m
• ↣ Initial Velocity of Stone (u) = 0m/s
• ↣ Acceleration Value (A) = 10m/s²

⚘ Need To Calculate :

• ➳ Time taken by stone to reach on ground ?

⚘ Formula Used Here :

• ✪ h = ut + 1/2 gt² ✪

________________

☢ Where :

• ⇒ h = Height
• ⇒ u = Initial Velocity
• ⇒ t = Time
• ⇒ g = Acceleration

★ Putting Values in Formula :

• ➬ h = ut + 1/2 gt²
• ➬ 180m = 0m/s × t + 1/2 × 10m/s² × t²
• ➬ 180 = 0 + 5 × t²
• ➬ 180 = 5t²
• ➬ 180/5 = t²
• ➬ t² = 36
• ➬ t = √36
• ➬ Time = 6 seconds

◆ Therefore :

• ❝ Time taken by stone to reach on ground is 6 Sec ❞

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Answer 2

Information provided with us:

• A stone is dropped from top of a tower of height 180 m
• g = 10 m/s²

What we have to calculate:

• We have to calculate and find out the time which had been taken by the stone to reach on the ground.

Using Formula:

Second equation of motion:-

• $\red{ \underline{ \boxed{\sf{S \: = \: ut {}^{2} + \dfrac{1}{2} at {}^{2} }}}}$

Where,

• S is the displacement caused
• u is the initial velocity
• a is acceleration
• t is time taken

As we know that initial velocity is always 0.

Values that we have,

• S is 180
• u is 0
• a is 10

Substituting the required values,

$: \longmapsto \: \tt{180 \: = \: 0 + \dfrac{1}{2} (10)(t) {}^{2} }$

$: \longmapsto \: \tt{180 \: = \: 0 + \dfrac{1}{2} \times (10)(t) {}^{2} }$

$: \longmapsto \: \tt{180 \: = \: 0 + \dfrac{1}{2} \times (10) \times (t) {}^{2} }$

$: \longmapsto \: \tt{180 \: = \: 0 + \dfrac{1}{2} \times 10 \times t {}^{2} }$

$: \longmapsto \: \tt{180 \: = \: 0 + \dfrac{1}{ \cancel2} \times \cancel10 \times t {}^{2} }$

$: \longmapsto \: \tt{180 \: = \: 0 + 1 \times 5 \times t {}^{2} }$

$: \longmapsto \: \tt{180 \: = \: 1 \times 5 \times t {}^{2} }$

$: \longmapsto \: \tt{180 \: = \: 5 \times t {}^{2} }$

$: \longmapsto \: \tt{t {}^{2} \: = \: \dfrac{180}{5} }$

$: \longmapsto \: \tt{t {}^{2} \: = \: \cancel\dfrac{180}{5} }$

$: \longmapsto \: \tt{t {}^{2} \: = \: 36 }$

$: \longmapsto \: \tt{t \: = \: \sqrt{36 } }$

$: \longmapsto \: \blue{\underline{\boxed{\bf{t \: = \: 6s}}}}$

$\small{ \underline{\bf{Hence, \: time \: taken \: by \: the \: stone \: to \: reach \: \: on \: the \: ground \: is \: of \: 6 \: seconds}}}$

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