Question

A BALL IS GENTLY DROPPED FROM A HEIGHT OF 20M. IF ITS VELOCITY INCREASES UNIFORMLY AT RATE OF 10M/S , WITH WHAT VELOCITY WILL IT STRIKE THE GROUND? AFTER WHAT TIME WILL IT STRIKE THE GROND?

Answer 1

s = 20m
u = 0m/s
a = 10m/s^2
 
Using third law of motion
v^2 = u^2+2as
v^2 = 0^2 + 2(10)(20)
v^2 = 400
v = 20m/s
 
Also, using first law of motion
v = u + at
20 = 0+10t
t = 2 sec

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