Following is an question from dc pandey mechanics part 1-introductory 1 Q-1)A particle is projected from ground with velocity 40sqrt2m/s.find- a)velocity and b)displacement of particle after 2sec .Take g=10m/s^2

given velocity v=40 root 2

v=u+at where a=g

1.v=u+at====>

we have to get the value by using this formula

2.s=ut+1/2gt^2 by using this formula wehave to get the value of displacements

thanks for asking these questions sorry idon''t have sufficient time to write the solution that''s why i wrote only formula what i know for that problem....

By using v=u+at we can find velocity at 2sec by taking a=g=-10m/s^2(a is negative because body moving upward but g acts in down ward direction)

v=40sqrt2-10*2=36.56m/sec.

b)Displacement can be find by using formula s=ut+(1/2)at^2 taking a=-10(same region explain above)

s=40sqrt2*2-(1/2)*10*4=93.13meters

velocity

v=u+at

a=-g

v=40√2-10*2

v=20(2√2-1)m/s

s=ut+at²/2

s=40√2*2-10*2*2

s=80√2-40

s=40(2√2-1)m

Question is hard

.v=u+at

what is the angle??????

without it how can u calculate the components of velocityy

after 2sec velocity=26.568 and displacement=33.136

velocity v=u+at a=-g v=40v2-10*2 v=20(2v2-1)m/s s=ut+at2/2 s=40v2*2-10*2*2 s=80v2-40 s=40(2v2-1)m