A balloon was moving upwards with a uniform velocity of 10 m/s. An object of finite mass is dropped from the balloon when it was at a height of 75 m from the ground level. The height of the balloon from the ground, when object strikes the ground, was around

(Take, the value of $g=10m/s^2$)

Given, initial speed of balloon, $u_b=10m{s}^{-1}$ 

Initial height of balloon from ground, h = 75m

initial speed of object, $u_a=-10m{s}^{-1}$

Let the time taken by object to reach ground be t.

$Ash=ut+\frac{1}{2}g{t}^2$

For object a,

$\begin{array}{l}\Rightarrow 75=-10t+\frac{1}{2}X10X{t}^2\\ \Rightarrow t^2-2t-15=0\Rightarrow t^2-5t+3t-15=0\\ \Rightarrow t(t-5)+3(t-5)=0\Rightarrow 5,-3\\ \therefore t=5s(\because t\ne -ve)\\ Also{u}_b=\frac{s}{t}\end{array}$

Where s is distance covered by balloon and t is time taken by object to reach ground

$\begin{array}{l}\therefore \text{}finalheight,h_f=h+s\\ =h+u_{b}Xt=75+10X5=75+50=125m\end{array}$