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Explanation / Answer

a)

at time 't' the distance in horizontal direction between the car and balloon=0

the distance traveelled by the car=Vot

so,

d-Vot=0

d=Vot

t=d/Vo ----(1)

time taken for the balloon to come down=t=sqrt(2h/g) --(2)

from (1) and (2)

d/V0=sqrt(2h/g)

Vo=d/sqrt(2h/g)

speed of the car,Vo= d/sqrt(2h/g)

b)

distance travelled by the car=(Vo*tb)+(Vo^2/as)-1/2*(as)*(Vo/as)^2

d-((Vo*tb)+(Vo^2/as)-1/2*(as)*(Vo/as)^2)=0

d-Vo*tb-Vo^2/as=0

(d-Vo^2/as)/Vo=tb

Vo-t*as=0 ===> t=Vo/as--(1)

t= time for which the car moves after the brakes are applied

time taken by the ballon to reach=sqrt(2h/g)

total time taken by the car=t+tb

t+tb=sqrt(2h/g)

from (1) )

Vo/as +tb =sqrt(2h/g)

Vo=as*(sqrt(2h/g) -tb)

c)substituting the value of tb in the above equation we get

Vo=as(sqrt(2h/g)-(d-Vo^2/as)/Vo)

=as(sqrt(2h/g)) -(d*as-Vo^2)/Vo

and solving for Vo

we get

Vo =d/sqrt(2h/g) -some thing

so Vo in part(2) is less than Vo in part(1)

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