Question

A thief throws a ring down at his partner at 4.0 m/s. It is 9.0 m to the partner’s hands.

a) Find the velocity of the ring when the partner catches it.

b) How long will it take for the ring to pass between the two crooks?

Answer 1

a) Vf = 13.87[m/s]

b) t = 1 [s]

Step-by-step explanation:

a)

To solve this problem we can use kinematics equations:

`v_(f)^(2)= v_(i)^(2)+(2*g*h)`

where:

Vf = final velocity [m/s]

Vi = initial velocity = 4 [m/s]

h = elevation = 9 [m]

g = gravity acceleration = 9.81 [m/s^2]

`v_(f)=\sqrt{(4)^(2) +(2*9.81*9)}\\v_(f)=13.87[m/s]`

b)

Now using the next equation we can find the time

`v_(f)=v_(i) +(g*t) \\t = (13.87-4)/9.81\\t=1[s]`