You don't say what the mass of the stone is, unless you are hinting that it is 1kg. (If you take g=10)
In any case, its initial PE is 10mg=100m, and its initial KE is (1/2)(m)(10^2) = 50m
So, at the bottom, its energy is all KE, in the amount of 150m = 1/2 m v^2, so v = √300
If stone is thrown directly down ward from the top of a 40m tall building with an initial speed of 10m/s,what will be the speed of the stone,when it reach the ground?g 10N/kg
4 answers
v = Vi + a t
Vi = -10
a = -9.81
v = -10 - 9.81 t
h = Hi + Vi t + (1/2) a t^2
0 = 40 - 10 t - 4.9 t^2
4.9 t^2 + 10 t - 40 = 0
https://www.mathsisfun.com/quadratic-equation-solver.html
t = 2 seconds
so
v = -10 - 9.81(2)
= - 29.6
Vi = -10
a = -9.81
v = -10 - 9.81 t
h = Hi + Vi t + (1/2) a t^2
0 = 40 - 10 t - 4.9 t^2
4.9 t^2 + 10 t - 40 = 0
https://www.mathsisfun.com/quadratic-equation-solver.html
t = 2 seconds
so
v = -10 - 9.81(2)
= - 29.6
V^2 = Vo^2+2g*h = 10^2+20*40 = 900
V = 30 m/s.
V = 30 m/s.
Firdt you answer my question