A classical mechanics problem by SRIJAN Singh

Time to cover half the circumference of the circle is $t=\dfrac {5π}{5}=π$ sec.

Change in velocity is $5-(-5)=10$ m/s.

Hence the average acceleration in this time interval is $<f>=\boxed {\dfrac {10}{π}}$ m/s $^2$ .

Solution to this problem :

Acceleration of the baloon is $1.25$ m/s $^2$ upwards. Time elapsed is $8$ sec. So the stone is released from the baloon at a height of $\dfrac 12\times 1.25\times 8^2=40$ m.

The velocity of the stone was $1.25\times 8=10$ m/s upwards.

After that, the stone takes a free fall with acceleration equal to the acceleration due to gravity : $10$ m/s $^2$ downwards.

Hence, if it takes time $t$ sec. to reach ground, then

$-40=10t-\dfrac 12\times 10\times t^2$

$\implies t^2-2t-8=0\implies t=\boxed 4$ sec.