Question : A balloon is ascending at the rate of 10 m/s at a height of 15 m above the ground when a packet is dropped from the balloon. After how much time does it reach the ground? (Take g=10 m/s²)
Doubt by Jaskirat
Solution :
Initial velocity (u) =10 m/s
displacement (s) = -15
acceleration (a) = g = -10 m/s²
Using Second Equation of Motion
s=ut+½at²
-15=10t+½(-10)t²
-15=10t+-5t²
5t²-10t-15=0 [Spliting the middle term]
5t²-(15-5)t-15=0
5t²-15t+5t-15=0
5t(t-3)+5(t-3)=0
(5t+5)(t-3)=0
5t+5=0
5t=-5
t=-5/5
t=-1
but time can't be negative so t ≠ -1
(t-3)=0
t=3
so t=3sec.
Hence, the packet reaches the ground after an time interval of 3 seconds.