Question

PHYSICS HELP URGENT!!! While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 7.59 m/s. The stone subsequently falls to the ground, which is 18.9 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. I need help finding out the impact speed, please.

Answer 1

v = 20.69 m/s

Step-by-step explanation:

Given that,

Initial speed of a stone, u = 7.59 m/s

he stone subsequently falls to the ground, which is 18.9 m below the point where the stone leaves his hand, h = 18.9 m

We need to find the speed of the stone impact the ground. Let the speed be v. Using the equation of kinematics to find v. So,

`v^2-u^2=2as`

Here, a = g

So,

`v^2=2gs+u^2\\\\v=√(2gs+u^2) \\\\v=√(2* 9.81* 18.9+(7.59)^2) \\\\v=20.69\ m/s`

So, the speed of the stone impact the ground is 20.69 m/s.