Question

a boy drops a coin from the top of a tower 50 m high .find the time by the coin to reach the ground . take g =10 ms*2
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Answer 1

Given :

• Height of tower, h = 50 m
• Acceleration due to gravity, g = 10 m/s²
• Initial velocity, u = 0 [Dropped from rest]

To calculate :

• Time taken to reach the ground, i.e time of descent.

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Solution :

When body is dropped from height then time of descent or time to reach the ground surface is given by,

$\qquad \star \underline{\boxed{\bf { t_d = \sqrt{\dfrac{2h}{g}}}}}$

• h = maximum height (projected vertically)
• g = acceleration due to gravity

$\dashrightarrow\quad \bf{ t_d = \sqrt{\dfrac{2h}{g}} }$

$\dashrightarrow\quad \bf{ t_d = \sqrt{\dfrac{2(50)}{10}} \; s }$

$\dashrightarrow\quad \bf{ t_d = \sqrt{\dfrac{100}{10}} \; s}$

$\dashrightarrow\quad \bf{ t_d = \sqrt{10} \; s}$

$\dashrightarrow\quad\underline{ \underline{\bf{ t_d = 3.16\;s }}}$

∴ The time taken by the coin to reach the ground is 3.16 seconds.

Answer 2

Answer:

Given :-

• A boy drops a coin from the top of a tower 50 m high.

To Find :-

• What is the time by the coin to reach the ground.

Formula Used :-

$\clubsuit$ Second Equation Of Motion Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{h =\: ut + \dfrac{1}{2} gt^2}}}\: \: \: \bigstar$

where,

• h = Height
• u = Initial Velocity
• t = Time Taken
• g = Acceleration due to gravity

Solution :-

Given :

• Height = 50 m
• Acceleration due to gravity = 10 m/s²
• Initial Velocity = 0 m/s

According to the question by using the formula we get,

$\implies \sf \bold{\purple{h =\: ut + \dfrac{1}{2} gt^2}}$

$\implies \sf 50 =\: (0)t + \dfrac{1}{2} \times (10)t^2$

$\implies \sf 50 =\: (0 \times t) + \dfrac{1}{2} \times 10 \times t^2$

$\implies \sf 50 =\: 0 + \dfrac{1 \times 10}{2} \times t^2$

$\implies \sf 50 =\: 0 + \dfrac{10}{2} \times t^2$

$\implies \sf 50 =\: 0 + 5 \times t^2$

$\implies \sf 50 - 0 =\: 5t^2$

$\implies \sf 50 =\: 5t^2$

$\implies \sf \dfrac{50}{5} =\: t^2$

$\implies \sf 10 =\: t^2$

$\implies \sf \sqrt{10} =\: t$

$\implies \sf 3.16 =\: t$

$\implies \sf\bold{\red{t =\: 3.16\: seconds}}$

$\therefore$ The time by the coin to reach the ground is 3.16 seconds .

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