Question

A high-wire artist missteps and falls 9.2 m to the ground. What is her velocity upon landing (just before she strikes the ground)?

Answer 1

Below

Step-by-step explanation:

Step-by-step explanation:

Her POTENTIAL energy (mgh)

will be converted to KINETIC energy (1/2 mv^2)

so

mgh = 1/2 mv^2 divide both sides of the equation by m

gh = 1/2 v^2 solve for 'v'

v = sqrt ( 2 g h) = sqrt ( 2 * 9.81 * 9.2 ) = 13.4 m/s

Answer 2

Step-by-step explanation:

We can use the kinematic equation to find the velocity of the high-wire artist just before she strikes the ground:

vf^2 = vi^2 + 2ad

where vf is the final velocity (the velocity just before she strikes the ground), vi is the initial velocity (which we can assume is 0), a is the acceleration due to gravity (which is approximately 9.81 m/s^2), and d is the distance fallen (which is 9.2 m).

Plugging in the values, we get:

vf^2 = 0 + 2(9.81 m/s^2)(9.2 m)

Simplifying:

vf^2 = 180.24 m^2/s^2

Taking the square root of both sides:

vf = 13.43 m/s

Therefore, the velocity of the high-wire artist just before she strikes the ground is 13.43 m/s.