A stone is thrown vertically upward with a speed of 22.0 m/s?

"A stone is thrown vertically upward with a speed of 22.0 m/s. How fast is it moving when it is at a height of 13.0 m ?"

Take g = 9.81 m/s², and take all upward quantities to be positive.

Initial velocity, u = 22.0 m/s
Acceleration, a = -9.81 m/s²
Displacement. s = 13.0 m

v² = u² + 2as
v² = 22.0² + 2 × (-9.81) × 13.0 m²/s²
v = ±√(22.0² - 2 × 9.81 × 13.0) m/s
Speed of the stone when it is a height of 13.0 m. |v| = 15.1 m/s
(The direction of the stone may be going upward or downward.)


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"How much time is required to reach this height?"

Assume that the stone is going upward.
s = ut + (1/2)at²
13.0 = 22.0t + (1/2) × (-9.81) × t²
4.905t² - 22.0t + 13.0 = 0
t = [22.0 ± √(22.0² - 4 × 4.905 × 13.0)] / (2 × 4.905) s

Time taken, t = 0.70 s when the stone is going upward.
Time taken, t = 3.78 s when the stone is going downward.

Energy is conserved
1/2 m * 22^2 = m g *13 + 1/2 m * v^2
divide BS by m
1/2 * 484 = 9.8 * 13 + 1/2 v^2
multiply BS by 2
484 = 9.8 * 26 + v^2
v^2 = (484-9.8*26)
v = sqrt( 484 - 9.8 * 26) = 15.1 m/s


as change in v = acceleration * time
22-15.1 = 9.8 * t