A rock is thrown vertically at the edge of a cliff 20 m above a river with a speed of 7.5 m/s. Calculate velocity and the distance from the river of the rock at 1.25 seconds after it is thrown.

1. Calculate the velocity at 1.25 seconds:

v = u + at

v = 7.5 - 9.8 \cdot 1.25

v = -4.75 \, \text{m/s}

2. Calculate the displacement from the cliff top at 1.25 seconds:

s = ut + \frac{1}{2}at^2

s = 7.5 \cdot 1.25 + \frac{1}{2} \cdot (-9.8) \cdot (1.25)^2

s=-1.71875\,\text{m}

3. Calculate the distance from the river:

\text{Distance from the river}=20+(-1.78175)

\text{Distance from the river}=18.28125\,\text{m}

Therefore, the velocity of the rock at 1.25 seconds is -4.75 \, \text{m/s} and the distance from the river is 18.28125\,\text{m} .