Question

how to solve this [ch 3 class 11 physics]

Answer 1

Balloon was going up with velocity of 40 m/s for 5 seconds.
s = vt
= 40×5
= 200 m
After ball was dropped:-
s = -200 m
u = 40 m/s
a = -10 m/s²
s = ut + 1/2at²
-200 = 40t + 1/2×-10t²
-200 = 40t-5t²
5t²-40t-200 = 0
a = 5
b = -40
c = -200
D = b²-4ac
= (-40)²-4×5×-200
= 1600+4000
= 5600
For finding roots, formula = -b±√D/2a
(i) (-(-40)+√5600)/2(5)
= (40+√(2×2×2×2×2×5×5×7))/10
= (40+20√14)/10
= 40/10+20√14/10
t = 4+2√14
(ii) (-(-40)-√5600)/2(5)
= (40-√(2×2×2×2×2×5×5×7)/10
= (40-20√14)/10
= 40/10-20√14/10
= 4-2√14
= 4-7.483
t = -3.483
Time cannot be -ve
Therefore, 4-2√14 terminated
t = 4+2√14

v = u+at
= 40 + (-10)(4+2√14)
= 40-40-20√14
= -20√14 m/s
Velocity was acting downwards at the end.
Therefore, it is negative.
We can write it +ve too

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