If you drop a stone from height of 2.5m. what is the speed of the stone right before it hits the ground?

Since the stone was dropped from height, its initial velocity = 0 m/s

Using  v² = u² + 2gs.

Where g ≈ 10 m/s²,  u = initial velocity = 0 m/s, s = height from drop = 2.5 m

v² = u² + 2gs

v² = 0² + 2*10*2.5

v² = 0 + 50

v² = 50

v = √50

v ≈ 7.07 m/s

Hence velocity just before hitting the ground is ≈ 7.07 m/s 

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