Question

Find displacement of the particle in 0-5 s.
(1) 20 m
(2) 22 m
(3) 40 m
(4) 25 m
​

Answer 1

Answer:

Explanation:

Given, s=20m,v=0m/sec,a=9.8m/s 2

1) Initial velocity v 2 −u 1

=2as

⇒0−u 2

=2(9.8)(20)u 2

=393m/sec

⇒u=19.8m/sec

2) Final velocity

u=0m/sec

v=?

a=9.8m/s

2

s=20m

v=19.8m/s

3)S

1=?,t

1 =0.5sec,S

2=?t 2 =2.5sec

For S

1 =ut 1 + 21a(t 1 )

2=19.8×0.5+ 21 (−9.8)(0)

2 =8.675m

For S

2 =ut 2 + 21a(t 2 ) 2

=19.8×2.5+ 21 (−9.8)(2.5)

2 =18.875m

Distance travelled=18.875−8.675=10.2m

4)t=?,u=19.8,s=15m,a=−9.8m/s 2

⇒S=ut+ 21​ (a)(t 2 )

⇒4.9t 2−19.8t+15=0

⇒t= 2×4.919.8±9.8

​

=1.01s or 3.03s

∴ When going up, t=3.03s

When coming down, t=1.01s