Question

a ball is thrown vertically upwards and goes to a height of 19.6m find its initial velocity
Total time of journey and final velocity when it strikes the ground
plz answer fast​

Answer 1

In this method consider the motion of stone directly from initial point (A)and the final point(C) on the ground .

A - C motion

For velocity, at the extreme point = 0 m/s.

Initial velocity = ?

we know,

$\leadsto{\tt{\pink{v^2 = u^2-2gs}}}$

$\leadsto{\tt{\pink{-u^2 = -2gs}}}$

$\leadsto{\tt{\pink{u^2 = 2× 9.8×19.6}}}$

$\leadsto{\tt{\pink{u= 19.6m/s}}}$

NoW,

${\tt{\green{Velocity\: striking\: the\: ground,}}}$

${\tt{\green{\:initial \:velocity = 0m/s}}}$

$\leadsto{\tt{\blue{v^2_1= u^2_1 +2gs}}}$

$\leadsto{\tt{\blue{v^2_1 = u^2_1+2×9.8×19.6}}}$

$\leadsto{\tt{\blue{v^2_1 = 19.6m/s}}}$

${\tt{\red{Now\:,Time\: for \:upward\: motion,,}}}$

${\tt{\red{ V=u+at}}}$

${\tt{\red{19.6 = 9.8t}}}$

${\tt{\red{9.8t =19.6,te}}}$

${\tt{\red{or,\: t\: = 2s}}}$

${\tt{\red{NoW, \: Time \:for\: downward\: motion,}}}$

$\leadsto{\tt{\blue{v^2_1 = u^2_1+ at_1}}}$

$\leadsto{\tt{\blue{19.6= at_1}}}$

$\leadsto{\tt{\blue{t= \frac{19.6}{9.8} = 2s}}}$

$\leadsto{\tt{\blue{Total\: time = (t_1+t_2) =2+2 =4s}}}$

Answer 2

QUESTION ....

ᴀ ʙᴀʟʟ ɪs ᴛʜʀᴏᴡɴ ᴠᴇʀᴛɪᴄᴀʟʟʏ ᴜᴘᴡᴀʀᴅs ᴀɴᴅ ɢᴏᴇs ᴛᴏ ᴀ ʜᴇɪɢʜᴛ ᴏғ 19.6ᴍ ғɪɴᴅ ɪᴛs ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ

ᴛᴏᴛᴀʟ ᴛɪᴍᴇ ᴏғ ᴊᴏᴜʀɴᴇʏ ᴀɴᴅ ғɪɴᴀʟ ᴠᴇʟᴏᴄɪᴛʏ ᴡʜᴇɴ ɪᴛ sᴛʀɪᴋᴇs ᴛʜᴇ ɢʀᴏᴜɴᴅ

ᴘʟᴢ ᴀɴsᴡᴇʀ ғᴀsᴛ

ANSWER.....

= ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ ,ᴜ = ?

= ᴡʜᴇɴ ʙᴀʟʟ ᴛʜʀᴏᴡɴ ᴠᴇʀᴛɪᴄᴀʟʟʏ ᴜᴘᴡᴀʀᴅs

= ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ = -g = -9.8ᴍ/${s}^{2}$

ᴀᴛ ᴇɴᴅ ᴘᴏɪɴᴛ ᴏғ ᴛʜᴇ ʙᴀʟʟ , ғɪɴᴀʟ ᴠᴇʟᴏᴄɪᴛʏ = 0m/s

= ${(v)}^{2} = {(u)}^{2}$ + 2as

> sᴏ, ${(0)}^{2} = {(u)}^{2}$ +2(-9.8)(19.6)

= ᴜ = 19.6m/s

(ii)

ғᴏʀ ɢᴏɪɴɢ ᴜᴘᴡᴀʀᴅ ᴅɪʀᴇᴄᴛɪᴏɴ

= v = u + at

= 0 = 19.6 + (9.8)(t)

= ᴛ = 2 sᴇᴄᴏɴᴅs

ᴛᴏᴛᴀʟ ᴛɪᴍᴇ = ᴜᴘᴡᴀʀᴅs + ᴅᴏᴡɴᴡᴏʀᴅs

= 2s + 2s = 4 sᴇᴄᴏɴᴅs

(iii)

ғᴏʀ ᴅᴏᴡɴᴡᴏʀᴅ ᴍᴏᴛɪᴏɴ

= U = 0m/s

= ᴠ = ? ( ᴅᴏɴᴛ ᴋɴᴏᴡ)

= ᴀ = +9.8ᴍ/s

= ᴛ = 2s

= ᴠ = ᴜ + ᴀᴛ

= 19.8 m/s

ʜᴇɴᴄᴇ ʏᴏᴜʀ ᴀɴsᴡᴇʀ ɪs 4 sᴇᴄᴏɴᴅs