Aperson 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)

-24.8 m/s

Step-by-step explanation:

Given:

y₀ = 60 m

y = 40 m

v₀ = 15 m/s

a = -9.8 m/s²

Find: v

There are three constant acceleration equations we can use:

y = y₀ + v₀ t + ½ at²

v = at + v₀

v² = 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/s²

v ≈ ±24.8 m/s

Since the coin is on the way down, the velocity is negative.  So v = -24.8 m/s.

where are the choices?

positive

step-by-step explanation:

the quadrant is facing upwards