A balloon,going upwards with a velocity of 12m/s,is at a height of 65m from the earth at any instant.Exactly at this instant a packet drops from it.How much time will the packet take in reaching the earth?(g=10m/s2)
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U = 12 m/s
s = -65 m
a = -10m/s²
s = ut + 1/2at²
-65 = 12t + 1/2×-10t²
-65 = 12t-5t²
5t²-12t-65 = 0
5t²+13t-25t-65 = 0
t(5t+13)-5(5t+13)
(t-5)(5t+13)
t = 5, t = -13/5
time cannot be-ve
Therefore, -13/5 terminated.
t = 5 seconds
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s = -65 m
a = -10m/s²
s = ut + 1/2at²
-65 = 12t + 1/2×-10t²
-65 = 12t-5t²
5t²-12t-65 = 0
5t²+13t-25t-65 = 0
t(5t+13)-5(5t+13)
(t-5)(5t+13)
t = 5, t = -13/5
time cannot be-ve
Therefore, -13/5 terminated.
t = 5 seconds
Hope it helps you ☺
Please mark as Brainliest ☺
Nyeshaverma:
thanks for ur help
welcome dear
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Answered by
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Sol. Initially the ball is going upward u = –7 m/s, s = 60 m, a = g = 10 m/s2 s = ut+ ½ at2 ⇒ 60 = -7t + ½ 10t2 ⇒ 5t2 – 7t – 60 = 0 t = (7+ √(49-4.5 (-60) ))/(2 x 5) = ( 7 ± 35.34 )/10 taking positive sign t = (7+ 35.34)/10 = 4.2 sec (∴ t ≠ -ve) Therefore, the ball will take 4.2 sec to reach the ground.
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