A ball is gently droped from a height of 20m .If its velocity increas uniformly at the rate of 10m per sec. With what velocity will its strick the ground? After what time will strike the ground
Answers
Answered by
4
given ,
height = 20m
acceleration = 10 m/s²
initial vel. =0 m/s
Final velocity = ?
Time = ?
from 3rd equation of motion ,
v² - u²= 2as
v²- 0² = 2×10×20
V² = 400 m
v =√400
v= 20 m/s
final velocity = 20m/s
from first equation of motion
v = u + at
20=0+10t
by transposing 10 to LHS
20/10= t
2 seconds = time
height = 20m
acceleration = 10 m/s²
initial vel. =0 m/s
Final velocity = ?
Time = ?
from 3rd equation of motion ,
v² - u²= 2as
v²- 0² = 2×10×20
V² = 400 m
v =√400
v= 20 m/s
final velocity = 20m/s
from first equation of motion
v = u + at
20=0+10t
by transposing 10 to LHS
20/10= t
2 seconds = time
Answered by
15
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