Question

A ball is gently dropped from a height of 20 .If its velocity increase uniformely at the rate of 10m s with what velociry will it strike the ground ? After what time will strike the ground

Answer 1

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

Answer 2

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ᴜ = 0 ( ᴀꜱ ᴛʜᴇ ᴏʙᴊᴇᴄᴛ ᴡᴀꜱ ɪɴɪᴛɪᴀʟʟʏ ᴀꜱ ʀᴇꜱᴛ)

ɢ = 10 ᴍ/ꜱ². (ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ ᴅᴜᴇ ᴛᴏ ɢʀᴀᴠɪᴛʏ)

ʜ = ꜱ = 20 ᴍ.

ᴡᴇ ʜᴀᴠᴇ ᴛᴏ ᴏʙᴛᴀɪɴ ᴛʜᴇ ᴛɪᴍᴇ .

ᴜꜱɪɴɢ ꜱᴇᴄᴏɴᴅ ᴇQᴜᴀᴛɪᴏɴ ᴏꜰ ᴍᴏᴛɪᴏɴ ꜰᴏʀ ꜰʀᴇᴇʟʏ ꜰᴀʟʟɪɴɢ ʙᴏᴅɪᴇꜱ

ꜱ = ᴜᴛ + 1/2 × ɢᴛ²

20 = 0 + 1/2 × 10 × ᴛ²

20 = 5ᴛ²

ᴛ² = 4

ᴛ = 2 ꜱ

ɴᴏᴡ, ᴜꜱɪɴɢ

ᴠ = ᴜ + ᴀᴛ

ᴠ = 0 + 10 × 2

ᴠ = 20 ᴍ/ꜱ.

ʜᴇɴᴄᴇ, ᴛɪᴍᴇ ᴡɪʟʟ ʙᴇ $\huge{\bold{\greeen{2sec}}}$ ᴀɴᴅ ᴛʜᴇ ʙᴀʟʟ ᴡɪʟʟ ꜱᴛʀɪᴋᴇ ᴡɪᴛʜ ᴀ ᴠᴇʟᴏᴄɪᴛʏ ᴏꜰ $\huge{\bold{\green{20m/s}}}$ .

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