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?
Answers
Answered by
31
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
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
Answered by
22
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
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