A ball is gently dropped from a height of 20 m. If it accelerates at the rate of 10 m/s^2, what will be its velocity when it strikes the ground?
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Answered by
2
v^2 - u^2 = 2gH
Initial Velocity (u) is zero
v^2 = 2gH
v = sqrt(2gH)
= sqrt(2 * 10 * 20)
= 20 m/s
Velocity when it strikes the ground is 20 m/s
Initial Velocity (u) is zero
v^2 = 2gH
v = sqrt(2gH)
= sqrt(2 * 10 * 20)
= 20 m/s
Velocity when it strikes the ground is 20 m/s
dandigeananya3outr8a:
Please tell me with another formula
To solve this type of questions you should know basic equations of motion.
v=u+at
v^2-u^2=2as
I used this formula
s=ut+1/2 at^2
Answer edited. Check it out.
which one
KTQ
You’re welcome :)
Answered by
16
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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