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A ball is dropped from a high rise platform at t=0 starting from rest . After 6 seconds another ball is thrown downwards from the same platform with a speed v . The two balls meet at t =18 s . What is the value of v.
Distance between the two balls after 6 secs = ½ * 10 * 62 = 180Relative velocity of second ball(wrt first ball) so that it can cover this distance = 180/ 12 = 15m/s Velocity of first ball (at t= 6) = 10* 6 = 60 v= 60 + 15 =75m/s
How come it's 180/12
The distance travelled by 1st ball in 18 sec is h=1/2gt^2=1/2*10*18^2=1620mNow to meet 2nd ball has to cover the same distance in (18-6 ) sec =12secSo for 2nd ballh=vt+1/2gt^21620=v*12+1/2*10*12^2=75m/s
diatance travelled by 1st ball in 18 sec=ut+1/2at^2s=0+1/2×10×18×18s=90×18s=1620mtime taken by second ball=18-6=12secso,distance travelled by second ball=ut+1/2gt×tso,1620==12×u+1/2×10×12×12u=135-60u=75m
Dear student,Please find the attached solution to your problem. since both cover the same distance downwards till t = 18hence, h = ½ g (18)2 = vt + ½ g t2where, t = 18 – 6 = 12shence, v * 12 + ½ g(12)2 = ½ g 182v = ½ g (182 – 122)/12v = 0.5 * 9.81 * (324 – 144)/ 12v = 73.575 m/s Hope it helps.Thanks and regards,Kushagra
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