a ball is dropped from a high rise platform at t=0. after 6sec another ball os thrown downwards from the same platform with the speed v.the two balls meet at t=18sec .what is the value of v? g=10m/s^2
Answers
Answered by
1
Answer:
Distance of 1st ball= S= 1/2 x 10 x 18
= 1620
Distance of 2nd ball = S' = ut + 1/2gt²
1620 = 12u + 5 x 144
u = 135 - 60
= 75 m/s
Similar questions