A ball is gently dropped from height of 20m. If it's velocity increases uniformly at the rate of 10meter per second square with what velocity will it strike the ground ?After what time it will strike the ground?
Answers
Answered by
6
here h=20m
acceleration=10m/s^2
initial velocity u=0
by equation of motion
h=ut+1/2at^2
20=0+1/2×10×t^2
t^2=4
t=2 seconds
acceleration=10m/s^2
initial velocity u=0
by equation of motion
h=ut+1/2at^2
20=0+1/2×10×t^2
t^2=4
t=2 seconds
Answered by
6
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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