A ball is dropped from rest from a height of 20.0 m. One second later a second ball is thrown vertically downwards. The two balls arrive on the ground at the same time. What is the initial velocity of the second ball?

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Answer 1 · 2018-03-03T01:23:50

#"15 m/s"#

Time taken by first ball to reach the ground is

#T = sqrt((2H)/g) = sqrt(("2 × 20.0 m")/("10 m/s"^2)) = "2 s"#

Time taken by second ball to reach the ground#= "2 s" - "1 s" = "1 s"#

For second ball

#S = ut + 1/2at^2#

#"20.0 m" = (u × "1 s") + (1/2 × "10 m/s"^2 × ("1 s")^2)#

#"20 = u + 5"#

#"u = 15 m/s"#

[#color(red)("Note:")# I took #"g = 10 m/s"^2# instead of #"9.8 m/s"^2# to make the calculations easier.]