Question

34. A body is thrown up with a velocity
29.23 msl. The distance traveled in last
but one second of upward motion is
1) 14.7 m
2) 9.8 m
3) 4.9 m
4) 1 m​

Answer 1

Answer:

4.4.9

solution:

steps:

1.we will find time to cover upward motion (i.e last second)

we have initial velocity u= 29.93m/s

at Max height, final velocity, v=0

acceleration due to gravity, a=-9.8m/s²

Then , by Newton's third equation of motion

v=u+at

0=29.23m/s - 9.8*t

t=2.98 s

2. Displacement in n'th second:-

displacement in 'n' second - displacement in (n-1)sec

Snth = Sn- Sn-1

=[un+1/2an²]-[u{n-1}+1/2a{n-1}²]

=u+1/2a(2n-1)

3. Here,

u=29.23m/s

a=-9.8m/s²

n=2.98s

Therefore, distance covered in 'nth' second

snth=u+1/2a(2n-1)

=29.23-9.8/2(2* 2.98-1)

=29.23-24.30

4.9m(approximately)

Hence, Displacement in that last second of upward motion is 4.9m.

HOPE U GET THE ANSWER GUYS