A hot-air balloonist, rising vertically with a constant velocity of magnitude `20 m s^(-1)`, releases a sandbag at an instant when the balloon is `25 m` above the ground . Afere it is released, the sandbag is in free fall. Skerch `a_(y)-t, v_(y)-t`, and `y-t` graphs for motion, taking origin at ground.
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Explanation

Apply constant acceleration equations to the vertical motion of the saandbag.
`y=y_(0)+v_(0y)t+(1)/(2)a_(y)t^(2)` …(i)
`v_(y)=v_(0y)+a_(y)t` …(ii)
Take +y` upward, `a_(y)=-10 m s^(1)`. `y_(0)=25 m`. The initial velocity of the sandbag equals the velocity of the ballooon, so `v_(0y)=+20 m s^(-2)`.
When the balloon reaches the fround, `y=0`. At its maximum height the sandbag has, `v_(y)=0`.
At maximum hight `0=20-10 t rArr t=2 s`
Position of the sand bag at maximum hithgt using(i)
`y=25+20xx2-(1)/(2)xx2^(2)=45 m`
The maximum height is `45 m` above the ground. For observer at fround the sand bat will go up and reach to its maximum hight `y_(max)=45 m` and then stop momentary and then start moving and after another `2` s it passes the point of droping. The and bag will reach fround where `y=0`.
Using (i)
`0=25+20t-5t^(2)`
`t^(2)-4t-5=0`
or `(t-5)(t+1)=0= rArrt=5 s`
The graphs of `a_(y). v_(y)`, and `y` versus `t` are given in . Take `y=0` at the grund. `
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