Question

A man in a balloon rising vertically with an acceleration of $4.9\, m\, s^{-2}4$ releases a stone 2 seconds after the balloon is let go from the ground. The greatest height above the ground reached by the stone is (Take $g = 9.8 \,m \,s^{-2}$)

• A. 14.7 m
• B. 19.6 m
• C. 9.8 m
• D. 24.5 m

Answer 1

Answer: (A)

14.7 m

Answer 2

Here, $a = 4.9\, m\, s^{-2},\, t = 2\, s,\, u = 0$ $S = ut + \frac{1}{2}at^{2}$ $s = 0 +\frac{1}{2} \times4.9 \times\left(2\right)^{2} = 9.8\,m$ This is the height from where stone is dropped. Upward velocity of stone when released, $\upsilon = u + at = 0 + 4.9 \times 2 = 9.8\, m\, s^{-1}$ The stone will move up till its velocity become zero. From, $\upsilon^{2} - u^{2} = 2as$ $0 - \left(9.8\right)^{2} = 2\left(-9.8\right)s'\quad\therefore\quad s' = 4.9\,m$ Maximum height above the ground $= s+s' = 9.8\, m + 4.9\, m = 14.7$