Stone is thrown vertically upward with a speed of 22.0 m/s. a) how fast is it moving when it reaches a height of 12.3 m?
b)how much time is required to reach this height?

a) v = 15.6 m/s

b) 0.65 s are needed to reach a height of 12.3 m

Explanation:

The equations that describe the height and velocity of the stone are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where

y = height of the stone at time t

y0 = initial height

v0 = initial speed

t = time

g = acceleration due to gravity

b) First, let´s find the time at which the stone reaches a height of 12.3 m:

y = y0 + v0 · t + 1/2 · g · t²

12.3 m = 0 m + 22.0 m/s · t + 1/2 · (-9.8 m/s²) · t²    (y0 = 0 placing the center of the frame of reference at the point at which the stone is thrown.)

-4.9 m/s² · t² + 22.0 m/s · t - 12.3 m = 0

t = 0.65 s (when the stone goes upward) and t = 3.84 s ( when the stone returns downward) .

So, 0.65 s are needed to reach a height of 12.3 m

a) The velocity at that time will be:

v = v0 + g · t

v = 22.0 m/s - 9.8 m/s² · 0.65 s = 15.6 m/s

dang that balls went far

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)