A man in a balloon rising vertically with an acceleration of $5 m / s^{2}$ releases a ball $2$ seconds after the balloon is let go from the ground. The greatest height (in $m$ ) above the ground reached by the ball is: ( $g = 10 m / s^{2}$ )

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Solution

Formula used: $s = u t + \frac{1}{2} g t^{2}$ , $v = u + a t$
Height ascended by balloon in $2 s$ :
$s_{1} = \frac{1}{2} \times 5 \times (2)^{2} = 10 m$
Velocity of balloon after $2 s$ :
$v_{1} = 5 \times 2 = 10 m / s$
This will also be the velocity of ball when released from balloon.
So further maximum height ascended by ball
$s_{2} = \frac{(10)^{2}}{(2 \times 10)} = 5 m$
Total height:
$s = s_{1} + s_{2} = 10 + 5 = 15 m$

Final answer: 15 m.

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A man in a balloon rising vertically with an acceleration of 5 m/s2 releases a ball 2 seconds after the balloon is let go from the ground. The greatest height (in m) above the ground reached by the ball is: (g=10m/s2)

A man in a balloon rising vertically with an acceleration of $5 m / s^{2}$ releases a ball $2$ seconds after the balloon is let go from the ground. The greatest height (in $m$ ) above the ground reached by the ball is: ( $g = 10 m / s^{2}$ )