Question

A stone is vertically thrown upwards with an initial velocity of 40 m/s^1 calculate maximum height reached

Answer 1

Given:

Initial velocity (u) = 40 m/s

Final velocity (v) = 0 m/s

Accelration (a) = - 10 m/s

To Find:

Maximum height reached (h)

Answer:

$\sf From \ 3^{rd} \ equation \ of \ motion \: we \: have: \\ \boxed{ \bf{ {v}^{2} = {u}^{2} + 2gh}}$

By substituting values in the equation we get:

$\rm \implies {0}^{2} = {40}^{2} + 2( - 10)h \\ \\ \rm \implies 0 = 1600 - 20h \\ \\ \rm \implies 20h = 1600 \\ \\ \rm \implies h = \dfrac{1600}{20} \\ \\ \rm \implies h = 80 \: m$

$\therefore$ $\boxed{\mathfrak{Maximum \ height \ reached \ (h) = 80 \ m}}$

Answer 2

GIVEN :-

• A sᴛᴏɴᴇ ɪs ᴠᴇʀᴛɪᴄᴀʟʟʏ ᴛʜʀᴏᴡɴ ᴜᴘᴡᴀʀᴅs ᴡɪᴛʜ ᴀɴ ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ ᴏғ 40 ᴍ/s .

TO FIND :-

• Tʜᴇ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ ᴛʜᴀᴛ ᴛʜᴇ sᴛᴏɴᴇ ʀᴇᴀᴄʜᴇᴅ .

SOLUTION :-

☯︎ ᴡʜᴇɴ ᴛʜᴇ sᴛᴏɴᴇ ɪs ʀᴇᴀᴄʜᴇᴅ ᴀᴛ ᴛʜᴇ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ, ᴛʜᴇɴ ᴛʜᴇɪʀ ɪᴛs ᴠᴇʟᴏᴄɪᴛʏ ɪs "0 ᴍ/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

• $\bf\red{u}$ = 40 m/s

• $\bf\red{a}$ = g = -10 m/s²

➳ 0² - (40)² = 2 × (-10) × h

➳ 0 - 1600 = -20 × h

➳ -1600 = -20 × h

➳ h = 1600/20

➳ h = 80 m

$\huge\red\therefore$ Tʜᴇ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ ᴛʜᴀᴛ ᴛʜᴇ sᴛᴏɴᴇ ʀᴇᴀᴄʜᴇᴅ ɪs "80 ᴍ" .