Question

.A stone is thrown vertically up wards with an initial velocity of 40 m/s taking g=10

m/s ^2.Find the maximum height reached by the Stone. What s the net displacement

and the total distance covered by the Stone?​

Answer 1

Answer:

Initial velocity = 40 m/s.

Acceleration due to gravity = -10 m/s²

Using the Formula :

⇒ v² - u² = 2ah

⇒ 0 - 40² = 2×(-10)×h

⇒ -1600 = -20h

⇒ h = -1600/-20

⇒ h = 80 m

So, the maximum height reached by the stone = 80 m.

Hence, the total distance travelled = 80 + 80 = 160 m.

And, the Displacement = 0 m.

Important Formulas :

1. v = u + at
2. s = ut + ½at²
3. v² - u² = 2as
4. Average speed = Total distance/Total time
5. Instantaneous velocity = dx/dt
6. Instantaneous acceleration = dv/dt
7. Average acceleration = ∆v/∆t
8. ∆x = x2 - x1 (Displacement)

Answer 2

GIVEN :-

• A sᴛᴏɴᴇ ɪs ᴛʜʀᴏᴡɴ ᴠᴇʀᴛɪᴄᴀʟʟʏ ᴜᴘᴡᴀʀᴅ .

• Tʜᴇ ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ ᴏғ ᴛʜᴇ sᴛᴏɴᴇ = 40m/s .

• g = 10m/s² .

TO FIND :-

1. Fɪɴᴅ ᴛʜᴇ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ ʀᴇᴀᴄʜᴇᴅ ʙʏ ᴛʜᴇ sᴛᴏɴᴇ .
2. Tʜᴇ ɴᴇᴛ ᴅɪsᴘʟᴀᴄᴇᴍᴇɴᴛ ᴏғ ᴛʜᴇ sᴛᴏɴᴇ .
3. Tʜᴇ ᴛᴏᴛᴀʟ ᴅɪsᴛᴀɴᴄᴇ ᴄᴏᴠᴇʀᴇᴅ ʙʏ sᴛᴏɴᴇ .

SOLUTION :-

$\huge\orange\star$ $\bf\pink{v^2\:-\:u^2\:=\:2as\:}$

Wʜᴇʀᴇ,

• ᴠ = ғɪɴᴀʟ ᴠᴇʟᴏᴄɪᴛʏ

• ᴜ = ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ

• a = ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ

• s = ʜᴇɪɢʜᴛ

Aᴄᴄᴏʀᴅɪɴɢ ᴛᴏ ᴛʜᴇ ǫᴜᴇsᴛɪᴏɴ,

• ᴠ = 0ᴍ/s

• ᴜ = 40ᴍ/s

• a = -10ᴍ/s²

• s = h = ?

➪ 0² - (40)² = 2 × -10 × h

➪ 0 - 1600 = -20 × h

➪ -1600 = -20 × h

➪ h = 1600/20

➪ h = 80m

$\huge\red{\therefore}$ ᴛʜᴇ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ ʀᴇᴀᴄʜᴇᴅ ʙʏ ᴛʜᴇ sᴛᴏɴᴇ is "80m" .

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$\huge\purple\checkmark$ Tᴏᴛᴀʟ ᴅɪsᴘʟᴀᴄᴇᴍᴇɴᴛ = 80 - 80

➪ Tᴏᴛᴀʟ ᴅɪsᴘʟᴀᴄᴇᴍᴇɴᴛ = 0ᴍ

$\huge\red{\therefore}$ Tʜᴇ ɴᴇᴛ ᴅɪsᴘʟᴀᴄᴇᴍᴇɴᴛ ᴏғ ᴛʜᴇ sᴛᴏɴᴇ ɪs "0ᴍ" .

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$\huge\purple\checkmark$ Tᴏᴛᴀʟ ᴅɪsᴛᴀɴᴄᴇ = 80 + 80

➪ Tᴏᴛᴀʟ ᴅɪsᴛᴀɴᴄᴇ = 160ᴍ

$\huge\red{\therefore}$ Tʜᴇ ᴛᴏᴛᴀʟ ᴅɪsᴛᴀɴᴄᴇ ᴄᴏᴠᴇʀᴇᴅ ʙʏ sᴛᴏɴᴇ ɪs "160ᴍ" .

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