Question

A balloon rises from ground with an acceleration of $1.25\, m s^{-2}$. After $8s$ a stone is released from balloon. The stone will

• A. cover a distance of 45 m in reaching the ground
• B. have a displacement of 50 m
• C. begin to move down after being released
• D. reach the ground in 4 s

Answer 1

Answer: (D)

reach the ground in 4 s

Answer 2

S = u + $\frac{1}{2} at^2 = \frac{1}{2} \times 1.25 \times 8^2$ = 40 m
Using V = u + at we get $\nu$ = 1.25 $\times$ 8 = 10 m $s^{-1}$
Using S = ut + $\frac{1}{2} at^2$ for dropped stone, we get
40 = -10t $\times + \frac{1}{2} \times 10t^2$ i.e. $5t^2$ - 10t - 40 = 0 i.e. t = 4 s or -2 s
Taking the possible value t = 4 s .