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

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