Question

the stone is dropped from height 20m . it velocity increases uniformly at 10m/s^2 during the fall.after how much time it will strike the ground and with velocity it will strike the ground

Answer 1

Given that,

Distance(s)="20m"

Initial velocity(u)=0m/s

Accelaration="10m/s-2"

Using the relation Distance-Time

s=v2 - u2=2a

s= v2-0=2*10

20=v2-0=20

v2-0=20*20

v2=400(underoot)

v=20 m/s

time=v=u=at

20=0+10

20 upon 10=0+10(t)=

time= 2 secs

Answer 2

ɢɪᴠᴇɴ

ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ(ᴜ) = 0 ᴍ/ꜱ

ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ (ᴀ) = 10 ᴍ/ꜱ²

ᴛɪᴍᴇ(ꜱ) = 20 ᴍ

ᴡᴇ ᴋɴᴏᴡ

ᴠ² = ᴜ² + 2ᴀꜱ

ᴘᴜᴛᴛɪɴɢ ᴠᴀʟᴜᴇꜱ:

ᴠ² = 0² + 2 (10 x 20)

ᴠ² = 400

ᴠ = 20 ᴍ/ꜱ

ꜰᴏʀ ᴛɪᴍᴇ:

ᴠ = ᴜ + ᴀᴛ

ᴛ = ᴠ - ᴜ/ᴀ

ᴛ = (20-0)10

  = 20/10

  = 2

∴ ꜱᴛʀɪᴋɪɴɢ ᴠᴇʟᴏᴄɪᴛʏ = 20 ᴍ/ꜱ

∴ ꜱᴛʀɪᴋɪɴɢ ᴛɪᴍᴇ = 2 ꜱᴇᴄᴏɴᴅꜱ