Question

A body of mass 50 kg falling from a certain height is

brought to rest after striking the ground with a speed of 5
m/s. If the resistance force of the ground is 500 N find
the duration of the contract
​

Answer 1

t = 0.5 seconds

Step-by-step explanation:

Given that,

Mass of a body, m = 50 kg

Initial speed, u = 0

Final speed, v = 5 m/s

The resistance force of the ground = 500 N

We need to find the time of contact. The formula for the force acting on an object is given by :

F = ma

a is acceleration

`F=(m(v-u))/(t)\\\\t=(m(v-u))/(F)\\\\t=(50* (5-0))/(500)\\\\t=0.5\ s`

So, the time of contact is 0.5 seconds.