A ball is gently dropped from height of 20m. If its velocity increases uniformly at the rate of 10m s-2,with what velocity will it strike the ground? After what time will it strike the ground?
Answers
Answered by
5
given, u=0
a= 10m/s-2
s=20m
s=ut+1/2at^2
20 = 0 × t + 1/2 × 10× t^2
20=5t^2
20÷5=t^2
4=t^2
t=2
hence , time =2s
v=u+at
v=0+10×2
v=20m/s
a= 10m/s-2
s=20m
s=ut+1/2at^2
20 = 0 × t + 1/2 × 10× t^2
20=5t^2
20÷5=t^2
4=t^2
t=2
hence , time =2s
v=u+at
v=0+10×2
v=20m/s
Tanishvishalbadgujar:
thanks for answer
Answered by
5
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
Similar questions