Question

A stone of mass 0.55 kilograms is released and falls to the ground. Measurements show that the stone has a kinetic energy of 9.8 joules at the point of impact with the ground. What is the stone's velocity when it hits the ground? A. 6.0 meters/second B. 7.2 meters/second C. 8.1 meters/second D. 9.3 meters/second

Answer 1

ke = 1/2 mv^2=9.8

sqrt(2x9.8/0.55)=v

sqrt(19.6/0.55 )

A 6.0

Answer 2

Answer:

A. 6.0 meters/second

Explanation:

As we know that the kinetic energy is given by

$KE = \frac{1}{2} mv^2$

now we know that

KE = 9.8 J

mass = 0.55 kg

now from above formula

$9.8 = \frac{1}{2}(0.55)v^2$

now by rearranging the values we have

$v^2 = \frac{2\times 9.8}{0.55}$

$v = 5.96 m/s$

So correct answer will be

A. 6.0 meters/second