A stone is dropped from a balloon going up with a uniform velocity of 5.0 m/s. If the balloon was 50 m high when the stone was dropped, find its height when the stone hits the ground. Take `g=10 m/s^2`.

Explanation

At t=0, the stone was going up with a velociyt of 5.0 m/sgt After that it moved as a freely falling particle with downward acceleration g. Take vertically upward as the positive X-axis. If it reaches the ground at time t,
`x=-50 m, u=5m/s, a=-10 m/s^2`.
we have `x=ut+1/2 at^2`
or `-50 m= (5m/s).t+1/2xx(-10m/s^2)t^2`
or, `t= (1+-sqrt(41))/2`s,
or, `t-2.7s or, 3.7s`
Negative t has no significance in this problem. The stone reaches the ground at t=3.7 s. During this time the balloon has moved uniformly up. The distance covered by it is
`5m/sxx3.7s=18.5m`
Hence, the height of the balloon whenthe stone reaches the ground is 50 m+18.5m=68.5m`.