A body ‘A’ is projected upwards with a velocity of 98 ms−1 and a second body B is projected upwards with the same initial velocity but after ‘4’ seconds. Both the bodies will meet after
A. 6 seconds
B. 8 seconds
C. 10 seconds
D. 12 seconds
Answers
- Velocity of body A = 98 m/s
- Body B projected upwards with the same initial velocity
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♦ Since they have same initial velocity , so their heights will be equal , H_A = H_B . As we know that ,
H = ut - 1/2 gt²
Let the time taken by the body A to reach point A be t . And here , the body B thrown after 4s , so the time taken will be ( t - 4 ) .
• H_A = H_B
⇒ ut - 1/2 gt² = u(t - 4) - 1/2 g (t - 4)²
⇒ ut - 1/2 gt² = ut - 4u - 1/2 ×g [t²-2(t)(4)+4²]
⇒ ut - 1/2gt² = ut - 4u - 1/2 g[t²-8t + 16]
⇒ ut - 1/2gt² = ut - 4u - 1/2 gt² - 1/2(-8gt) - 1/2(16g)
⇒ ut - 1/2gt² - ut + 1/2gt² = - 4u + 4gt + 8g
⇒ 0 = -4u + 4gt - 8g
⇒ 4u - 4gt + 8g = 0
⇒ 4(98) - 4(9.8)t + 8(9.8) = 0
⇒ 392 - 39.2t + 78.4 = 0
⇒ 470.4 = 39.2 t
⇒ t = 12s
Bodies will meet after 12s .
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Answer
ᴛʜᴇ ᴅɪsᴛᴀɴᴄᴇ ʙᴇᴛᴡᴇᴇɴ ᴛʜᴇ ᴡʀɪᴛᴛᴇɴ ᴛᴇxᴛ ᴀɴᴅ ᴛʜᴇ ᴇᴅɢᴇ ᴏғ ᴛʜᴇ ᴘᴀᴘᴇʀ ɪs ᴄᴀʟʟᴇᴅ ᴍᴀʀɢɪɴ. ᴛʜᴇʀᴇ ᴀʀᴇ 4 ᴛʏᴘᴇs ᴏғ ᴍᴀʀɢɪɴs ᴘʀᴇsᴇɴᴛ ɪɴ ᴍɪᴄʀᴏsᴏғᴛ ᴡᴏʀᴅ :-
ᴛᴏᴘ ᴍᴀʀɢɪɴ :- ᴅɪsᴛᴀɴᴄᴇ ʙᴇᴛᴡᴇᴇɴ ᴛʜᴇ ᴛᴇxᴛ ᴀɴᴅ ᴛʜᴇ ᴛᴏᴘ ᴇᴅɢᴇ ᴏғ ᴛʜᴇ ᴘᴀɢᴇ.
ʙᴏᴛᴛᴏᴍ ᴍᴀʀɢɪɴ :- ᴅɪsᴛᴀɴᴄᴇ ʙᴇᴛᴡᴇᴇɴ ᴛʜᴇ ᴛᴇxᴛ ᴀɴᴅ ᴛʜᴇ ʙᴏᴛᴛᴏᴍ ᴇᴅɢᴇ ᴏғ ᴛʜᴇ ᴘᴀɢᴇ.
ʟᴇғᴛ ᴍᴀʀɢɪɴ :- ᴅɪsᴛᴀɴᴄᴇ ʙᴇᴛᴡᴇᴇɴ ᴛʜᴇ ᴛᴇxᴛ ᴀɴᴅ ᴛʜᴇ ʟᴇғᴛ ᴇᴅɢᴇ ᴏғ ᴛʜᴇ ᴘᴀɢᴇ.
ʀɪɢʜᴛ ᴍᴀʀɢɪɴ :- ᴅɪsᴛᴀɴᴄᴇ ʙᴇᴛᴡᴇᴇɴ ᴛʜᴇ ᴛᴇxᴛ ᴀɴᴅ ᴛʜᴇ ʀɪɢʜᴛ ᴇᴅɢᴇ ᴏғ ᴛʜᴇ ᴘᴀɢᴇ.