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Question

(A) The balloon starts from rest with acceleration $a = 4.9 \, m/s^2$. After $t = 2 \, s$,the velocity of the balloon (and the ball) is $v = u + at =

Vedclass · Accepted Answer

$14.7$

Vedclass · Answer

(A) The balloon starts from rest with acceleration $a = 4.9 \, m/s^2$. After $t = 2 \, s$,the velocity of the balloon (and the ball) is $v = u + at = 0 + 4.9 	imes 2 = 9.8 \, m/s$.The height of the balloon at $t = 2 \, s$ is $h = \frac{1}{2}at^2 = 0.5 	imes 4.9 	imes (2)^2 = 9.8 \, m$.When the ball is released,it has an initial upward velocity of $9.8 \, m/s$ at a height of $9.8 \, m$. The additional height $s$ reached by the ball after release is given by $v^2 = u^2 - 2gs$,where the final velocity at the peak is $0$.$0 = (9.8)^2 - 2 	imes 9.8 	imes s \implies s = \frac{9.8 	imes 9.8}{2 	imes 9.8} = 4.9 \, m$.The greatest height above the ground is $H = h + s = 9.8 + 4.9 = 14.7 \, m$.

$A$ 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 .......... $m$ :-

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