Question

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

Answer 1

We know that,
                     the initial velocity, u = 0
                     acceleration, a = 10 m/$s^{2}$     (Taking down as +ve)
                     distance, s = 20 m

Using Second equation of Motion,
     
                     s = ut + $\frac{1}{2}$a$t^{2}$
                   20 = 0 + 5$t^{2}$
                   => t = 2 s
Therefore, it will strike the ground after 2 seconds
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Now, using the 1st equation of Motion, we have

                                 v = u + at
                                 v = 0 + 10 x 2
                                 v = 20 m/s
Therefore the velocity at which it will strike the ground is 20 m/s

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