A man in a balloon rising vertically with an | vedclass

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(A) $1$. Height reached by the balloon in $t = 2\,s$:$h_1 = \frac{1}{2} a t^2 = \frac{1}{2} 	imes 4.9 	imes (2)^2 = 9.8\,m$.$2$. Velocity of the bal

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$14.7$

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(A) $1$. Height reached by the balloon in $t = 2\,s$:$h_1 = \frac{1}{2} a t^2 = \frac{1}{2} 	imes 4.9 	imes (2)^2 = 9.8\,m$.$2$. Velocity of the balloon at $t = 2\,s$:$v = a t = 4.9 	imes 2 = 9.8\,m/s$.$3$. When the ball is released,it has an initial upward velocity $u = 9.8\,m/s$. It moves under gravity until its velocity becomes zero at maximum height $h_2$ relative to the release point:$v^2 = u^2 - 2 g h_2 \implies 0 = (9.8)^2 - 2 	imes 9.8 	imes h_2 \implies h_2 = \frac{9.8}{2} = 4.9\,m$.$4$. The greatest height above the ground is the sum of the height at release and the additional height gained by the ball:$H = h_1 + h_2 = 9.8 + 4.9 = 14.7\,m$.

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

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