Answer on Question #60032-Physics-Mechanics-Relativity
A clown in circus juggles with n balls using only one hand. He throws each ball vertically upwards with the same speed V at equal time intervals T. Denote acceleration of free fall by g.
(a) Find expressions for the speed of projection and height of the i th ball above his hand when he throws the n th ball.
Let he uses n=4 balls and when he throws the fourth ball, the distance between the second and third ball is d=50 cm.
(b) Where is the first ball, when the juggler throws the fourth ball?
(c) What is maximum height attained by each ball above the hands of the juggler?
Solution
(a)
v=V−gt,h=Vt−2gt2.t=T(n−i).
The total time of flight is
τ=nT.
At this time:
v=V−gnT=−VV=2gnT
Therefore,
h=Vt−2gt2=2gnTT(n−i)−2gT2(n−i)2=2gT2i(n−i)
(b)
h(2)=2gT22(4−2)=2gT2h(3)=2gT23(4−3)=23gT2d=2gT2−23gT2=21gT2.h(1)=2gT21(4−1)=23gT2=3d=150 cm=1.5 m.
(c)
H=h(2τ)=h(2nT)=h(24T)=h(2T)=2g4T2T−2g(2T)2=2gT2H=4d=2m.
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