Question

A ball is gently dropped from a height 20m.If its velocity increases uniformly at the rate of 20m/s square.With what velocity will it strike the ground and after what time it will strike the ground?

Answer 1

Formula to be used
=================
S = ut +½at²
V = u + at

Where S = displacement
V= final velocity
u = initial velocity
a = Acceleration = change in velocity with respect to time
t = time taken.
__________________________

S =20 m
a = 20 m/s²
u = 0 m/s
v =?
t =?

By formula (1)__

S = ut +½×at²

or, 20 = 0 +½×20 ×t²
$so.. \: t = \sqrt{2} = 1.414 \: \: sec$
Now by formula (2)__

V = u +at

or, V =0 + 20 ×1.414

or, V = 28.28 m/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