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.

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer

answer: none of these answers are correct only the second half is

step-by-step explanation: