Question

a body is projected vertically upward with velocity 30m/s . the distance travelled by the body in 3rd second is??

Answer 1

use the formula of distance traveled in t second or you use the concept of Sn -Sn-1

Answer 2

A body is projected vertically upward with velocity 30m/s . we have to find the distance travelled by body in 3rd second.

first of all, we have to see when body becomes rest momentarily at maximum height.
take final velocity, v = 0,
initial velocity, u = 30 m/s
use formula , v = u + at
0 = 30 + g × t
0 = 30 + (-10)t
t = 3 sec.
hence, body reaches at maximum height after 3 sec. it means path of body during 0 to 3 sec is continuous.

use formula,
$S_{nth}=u+\frac{g}{2}(2n-1)$
now, distance travelled by body in 3rd second, $S_{3rd}=|u+\frac{g}{2}(2\times3-1)|$
= |30 + (-10)/2 × 5|
= |30 - 25|
= 5m