Question

the velocity of a freely falling body after t sec is 49m/sec .its velocity after t+2sec will be

Answer 1

v =  49 m/sec

u = 0 m/sec

a = g = 9.8 m/sec^2

v = u + at ...|| FIRST EQUATION OF MOTION

49 = 0 + (9.8)t

49 = (9.8)t

t = 5 sec

t + 2 = 7

v = u + at  ...|| FIRST EQUATION OF MOTION

v = 9.8*7 = 68.6 m/sec

Answer 2

Given:

• Velocity at t sec = 49 m/s

To find:

Velocity after t + 2 sec ?

Calculation:

Let's assume that the freely falling body was dropped from a height when the initial velocity can be considered zero.

So, we can say:

$v = u + gt$

$\implies \: 49= 0 + (9.8)t$

$\implies \: (9.8)t = 49$

$\implies \: t = 5 \: sec$

So, velocity after (5+2) = 7 sec will be :

$v_{2} = u + gt_{2}$

$\implies v_{2} = 0 + (9.8)7$

$\implies v_{2} = 68.6 \: m {s}^{ - 1}$

So, velocity after t+2 seconds is 68.6 m/s