Answer to Question #4153 in Mechanics | Relativity for Jenny

Question

A ball is thrown upward from the ground with an initial speed of 18.2 m/s; at the same instant a ball is dropped from rest from a building 12 m high. After how long will the balls be at the same height?

Answer 1

Let's write the equations of motion for both objects:
1. H1= H₁₀ + v₁₀t - gt²/2, H₁₀=0, v₁₀ = 18.2 m/s, g = 9.8 m/s²
2. H2 = H₂₀ + v₂₀t - gt²/2, H₂₀=12, v₂₀ = 0 m/s, g = 9.8 m/s²


Thus
H1 = H2
v₁₀t - gt²/2 = H₂₀ - gt²/2

v₁₀t = H₂₀

t = H₂₀/v₁₀ = 12/18.2 = 0.66 s