What is the final velocity, in m/s, of a hoop that rolls without slipping down a 8.20 m high hill, starting from rest?

12.68 m/s.

Explanation:

Equations of motion:

i. S = vi*t + 1/2 * a*(t^2)

ii. vf = vi + a*t

iii. vf^2 = vi^2 + 2a*S

Where, vf = final velocity

vi = initial velocity

S = distance travelled

a = acceleration due to gravity

t = time taken

Given:

vi = 0 m/s

S = 8.2 m

vf = ?

a = 9.81 m/s^2

Using iii. Equation of motion,

vf^2 = vi^2 + 2a*S

= 2 * 9.81 * 8.2

= 160.884

vf = sqrt (160.884)

= 12.68 m/s.

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle

answercollision;