Question

stone is thrown vertically upward with a speed 10ms from the edge of a cliff 65m high it's speed just before hitting the bottom

Answer 1

Dear STUDENT
s = displacement = -65m
V = final velocity what you want
u = initial velocity 10m/s up = +10
a = acceleration =G= 9.8m/s² down = -9.8
Equation:
V² = U² +2*a*s
V² = 10² +2*(-9.8)*(-65)
V² = 100 +1274
V² = 1374
V = 37.06 m/s
it's speed just before hitting the bottom will be 37.06 m/s​​​​
thank you :)