Question

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10ms2 , with what velocity will it strike the ground? After what time will it strike the ground?

Answer 1

We have ,
u= 0 m/s ( as u r just dropping the ball)
a = 10 ms^-2
h=20 m
( Here instead of distance we use height )
From the Third Equation of motion
$v {}^{2} - u {}^{2} = 2as$
$v { }^{2} = 2as + u {}^{2}$
v^2 = 2 × 10 × 20 + 0
v^2 = 400
v = root of 400
v = 20 m/s

From the First equation of motion ,
v = u + at
t = v - u / a
= 20 - 0/ 10
= 20/10
= 2 s

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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