Answer: time is the same
Explanation: the distance(H) is the same in each case .
we drop the balls , no drag force using basic kimnematics
y =gt*t/2 , yo=0 , vo=0 , y=H , so : t= sqrt(2H/g)
comment: if distance H starts to grow....we could begin to note a difference because of gravity g is smaller as we go up
Same time will be taken in both the cases
The newton's kinematic equations are a set of equations that describe the motion of an object with constant acceleration
Using newton's kinematic equation
s = ut + 1/2 g [tex]t^{2}[/tex]
u=0
t = [tex]\sqrt{2s/g}[/tex]
let height of tower = h
t1 (time taken by P to reach the middle of the tower) = [tex]\sqrt{(2* (h/2))/g}[/tex]
= [tex]\sqrt{h/g}[/tex]
t2 (time taken by Q to reach the ground) = [tex]\sqrt{(2* (h/2))/g}[/tex]
= [tex]\sqrt{h/g}[/tex]
t1 = t2
hence same time will be taken in both the cases
learn more about newton's kinematic equation
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