While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 7.19 m/s. the stone subsequently falls to the ground, which is 20.1 m below the point where the stone leaves his hand. at what speed does the stone impact the ground? ignore air resistance and use =9.81 m/s² for the acceleration due to gravity.

21.1 m/s

Explanation:

The motion of the stone is a uniformly accelerated motion (free fall), so we can find the final velocity of the stone by using the following suvat equation

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the vertical displacement

For the stone in this problem, taking upward as positive direction, we have:

u = +7.19 m/s is the initial velocity

is the acceleration of gravity

s = -20.1 m is the displacement

Solving for v, we find the final velocity:

And the correct solution is the one with negative sign, since the final velocity is downward:

Therefore, the final speed (the magnitude of the velocity) is

double the distance = divide strength by 4

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer