Question

1. A bag dropped from a helicopter falls with an acceleration of 9.8m/s2. What is its velocity after 5s?

Answer 1

The answer is 49m/s.

Step-by-step explanation:

Given,

acceleration (a) =9.8m/s^2

initial velocity (u)=0 (as it was in rest condition in helicopter)

time (t) =5 s

now, final velocity (v) after 5s =?

we have,

`a = (v - u)/(t)`

`or \: 9.8 = (v - 0)/(5)`

or, v = 49m/s

Therefore, the velocity after 5s is 49m/s.

Hope it helps..