a ball is dropped from a high rise platform t=0 starting from rest. After 6s another ball is thrown downwards from the same platform with a speed v.the two balls meet at t=18s .what is the value of v?(g=10m/s2)
Answer (1)
Student Expert
14th Sep, 2019
Hey!There
Both balls will meet at same distance,so equate with displacement covered by body..Use S=ut+1/2(at^2)..For first ball take t=18sec while t=12sec as ball is dropped after 6sec..And finally you will get u=75m/s..
Ask if any query.
2 Comments
Comments (2)
I get u =52.5m/s by your process sir.but answer is u=75m/s.
Reply
Tania verma
Sorry,I did mistake.as ball is dropped after 6sec so time taken 2nd ball to cover same distance will be 12sec..
