HELP PLEASE !!!!! A person is on the top of a 60 m tall building. They throw a coin up into the air with a velocity of 15 m/s. How fast is the coin moving when it has fallen 20 m away underneath the person? (Show all work)

Answer:-24.8 m/sStep-by-step explanation:Given:y₀ = 60 my = 40 mv₀ = 15 m/sa = -9.8 m/s²Find: vThere are three constant acceleration equations we can use:y = y₀ + v₀ t + ½ at²v = at + v₀v² = v₀² + 2a(y − y₀)We aren't given the time, so we need to use the third equation, which is independent of time:v² = v₀² + 2a(y − y₀)Plug in the values:v² = (15 m/s)² + 2(-9.8 m/s²) (40 m − 60 m)v² = 617 m²/s²v ≈ ±24.8 m/sSince the coin is on the way down, the velocity is negative.  So v = -24.8 m/s.