Referring to the previous Illustration, if the stone reaches the ground after 1 2 scoond from same initial height of release, find the (a) speed of the balloon af the time of releasing the stone, (b) total distance covered by the stone till it reaches The ground level, c) The average speed and (d) average velocity of the stone for the total time of its flight, one from the bottom and the other from the top.

Explanation

(a) 4.8 = `-v_(0)t+(1)/(2)gt^(2)`
Putting t= 2 sec, We obtain `2V_(0)=20-4.8=15.2m//s`
`rArrv_(0)=7.6m//sec`.
(b)`d=4.8+2xx(V_(0)^(2))/(2g)=4.8+(V_(0)^(2))/(g)`where `v_(0)=7.6m//sec`.
(c)(d)`vecU=(d)/(t)`,put the values of d and t = 2 sec.`vecv=(s)/(t)`put s=4.8 and2 sec. to obtain `vecv = 2.4 m//s`