Question

A ball is gently dropped from a height of 20m . if its velocity increases uniformly at a rate of 10m/s², with what velocity will it strike the ground? After what time will it strike the ground?

Answer 1

Initial velocity(u) = 0 m/s
Acceleration (a) = 10 m/s²
Time(s) = 20 m

We know,

v² = u² + 2as

Putting values:

v² = 0² + 2 (10 x 20)
v² = 400
v = 20 m/s

For time:
 v = u + at
t = v - u/a
t = (20-0)10
  = 20/10
  = 2

∴ Striking velocity = 20 m/s ← (Answer)
∴ Striking Time = 2 seconds  ← (Answer)

Good Studies!

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