Question

Consider the scenario where a person jumps off from the edge of a 1 m high platform and lands on the ground Suppose his initial jumping speed was 3 m/s. For how long was this person in the air?

Answer 1

For 0.24 sec the person was in the air.

Step-by-step explanation:

Given that,

Height = 1 m

Initial velocity = 3 m/s

We need to calculate the time

Using equation of motion

`s=ut+(1)/(2)gt^2`

Where, u = initial velocity

s = height

Put the value into the formula

`1 =3* t+(1)/(2)*9.8* t^2`

`4.9t^2+3t-1=0`

`t = 0.24\ sec`

On neglecting the negative value of time

Hence, For 0.24 sec the person was in the air.