Question

solve it........... ​

Answer 1

GIVEN :-

• A sᴛᴏɴᴇ ɪs ᴛʜʀᴏᴡɴ ᴠᴇʀᴛɪᴄᴀʟʟʏ ᴜᴘᴡᴀʀᴅ ᴡɪᴛʜ ᴀɴ ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ ᴏғ "14 m/s" .

TO FIND :-

• Tʜᴇ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ ᴛʜᴀᴛ ᴛʜᴇ sᴛᴏɴᴇ ʀᴇᴀᴄʜᴇᴅ .
• Tʜᴇ ᴛɪᴍᴇ ᴏғ ᴀsᴄᴇɴᴛ .

SOLUTION :-

• ☯︎ ᴡʜᴇɴ ᴛʜᴇ sᴛᴏɴᴇ ɪs ʀᴇᴀᴄʜᴇᴅ ᴀᴛ ᴛʜᴇ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ, ᴛʜᴇɴ ᴛʜᴇɪʀ ɪᴛs ᴠᴇʟᴏᴄɪᴛʏ ɪs "0 m/s" .

ᴡᴇ ʜᴀᴠᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

$\huge\red\star$ $\bf\purple{v^2\:-\:u^2\:=\:2\:a\:s\:}$

Wʜᴇʀᴇ,

$\bf\red{v}$ = ғɪɴᴀʟ ᴠᴇʟᴏᴄɪᴛʏ

$\bf\red{u}$ = ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ

$\bf\red{a}$ = ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ

$\bf\red{s}$ = ᴅɪsᴛᴀɴᴄᴇ/ʜᴇɪɢʜᴛ(h)

Aᴄᴄᴏʀᴅɪɴɢ ᴛᴏ ᴛʜᴇ ǫᴜᴇsᴛɪᴏɴ,

$\bf\red{v}$ = 0 m/s

$\bf\red{u}$ = 14 m/s

$\bf\red{a}$ = g = -9.8 m/s

⟹ 0² - (14)² = 2 × -9.8 × h

⟹ 0 - 196 = -19.6 × h

⟹ -196 = -19.6 × h

⟹ h = -196/-19.6

⟹ h = 10 m

$\huge\red\therefore$ Tʜᴇ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ ᴛʜᴀᴛ ᴛʜᴇ sᴛᴏɴᴇ ʀᴇᴀᴄʜᴇᴅ is "10 m/s" .

$\huge\blue\star$ $\bf\green{Time\:of\:ascent\:=\:\dfrac{u}{g}\:}$

• ➳ Tɪᴍᴇ ᴏғ ᴀsᴄᴇɴᴛ = 14/9.8

• ➳ Tɪᴍᴇ ᴏғ ᴀsᴄᴇɴᴛ = 1.428 s

$\huge\red\therefore$ Tʜᴇ ᴛɪᴍᴇ ᴏғ ᴀsᴄᴇɴᴛ ɪs "1.428 s" .

Answer 2

Answer:

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large\sf {Given :- }$

▪ A stone is thrown vertically upward with an initial velocity of u = 14 ms⁻¹

▪ Acceleration due to gravity, g = 9.8 ms⁻²

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large \sf{To \: Find :-}$

▪ Maximum height reached.

▪ Time of ascent.

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large \sf{Solution :-}$

Since this situation can be assumed as a free fall, so we can apply the simpler formulae for finding the maximum height and the time of ascent.

Although, this can also be solved by using the Newton's laws of motion.

Anyways, let's move forward to finding the answer.

Here, we have

u = 14 ms⁻¹

g = 9.8 ms⁻²

We know,

⇒ Maximum height = u² / 2g

⇒ Max. height = 14² / 2×9.8

⇒ Max. height = 196 / 19.6

⇒ Max. height = 10 m

So, The maximum height achieved by the stone is 10 m.

Now, the time of ascent can be calculated as,

⇒ Time of ascent = u / g

⇒ Time of ascent = 14 / 9.8

⇒ Time of ascent = 1.429 s

Hence, The time of ascent is 1.429 s.

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