Question

ᴀ ᴘᴀʀᴛɪᴄʟᴇ ɪs ᴅʀᴏᴘᴘᴇᴅ ᴜɴᴅᴇʀ ɢʀᴀᴠɪᴛʏ ғʀᴏᴍ ʀᴇsᴛ ᴀ ʜᴇɪɢʜᴛ h ᴀɴᴅ ɪᴛ ᴛʀᴀᴠᴇʟs ᴀ ᴅɪsᴛᴀɴᴄᴇ
$\tt\:\dfrac{9h}{25}$
ɪɴ ᴛʜᴇ ʟᴀsᴛ sᴇᴄᴏɴᴅ,ᴛʜᴇ ʜᴇɪɢʜᴛ h ɪs....?
$\tt\:[ take \: g = 9.8 \: ms^{-2} ]$
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Answer 1

Answer:

h = 122.5 m

Explanation:

A particle is dropped from a height h.

Then,

Initial velocity = 0

acceleration = a = g

distance traveled = h

time taken = t seconds

=> h = (1/2) gt^2

=> h = (g/2) t^2 ------ (1)

=> (16/25)h = (g/2) t^2

Now for the case when the object travels from the top to the point (t-1) now for the displacement :

=> (16/25) = (g/2)(t-1)^2]

Then it will become,

=> 16/25 = [(t-1)/t]^2

On taking Square root on both sides, we get

=> (4/5) = (t - 1)/t

=> t = 5

Place in (1), we get

=> h = (g/2) * t^2

=> h = (g/2) * 25

=> h = (9.8/2) * 25

=> h = 122.5 m

Hope it helps!

Answer 2

ANSWER:-

➣

$\bf{Height= 122.5m}$

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• Given:-

Distance in last second $\bf{D= {\frac{9h}{25}}}$

Let's consider that a particle takes time $\bf[t]$ to reach the ground.

• Using equation of motion:-

The total distance will be:-

$\boxed{\bf{s = ut + \frac{1}{2} \: g {t}^{2}}}$

$h = 0 +{\frac{1}{2}}\times {9.8} \times {t}^{2} \: \: \: \longrightarrow {\bf(eqn.1)}$

The particle covers the distance D' in [t-1] second.

$D' = h - D$

$D' = h - {\frac{9h}{25}}$

$D' = {\frac{16h}{25}}$

• Using equation of motion:-

${\frac{16h}{25}}= 0 + {\frac{1}{2}}g {t-1}^{2} \: \: \: \longrightarrow {\bf(eqn.2)}$

• Dividing equation [2] by equation [1] :-

${\frac{16}{25}}={\frac{{t-1}^2}{{t}^2}}$

${\frac{16}{25}}\times {t}^{2} = (t-1)^{2}$

${\frac{4}{5}}\times {t} = t-1$

$\bf{t= 5 sec.}$

• Putting value of t in eqn.1:-

$h={\frac{1}{2}}\times {9.8} \times {5} \times {5}$

$\bf{h= 122.5m}$

Therefore,

$\implies \large{The \: Height \: is \: {\bf{122.5}}}$

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