Question

a particle is projected with velocity 9m/s up a line of greatest slope of a smooth plane inclined at an angle a to the horizontal where sin a=4/7. find the speed of the particle when it has travelled 5m up the plane.

Answer 1

u get a by taking arcsin 4/7

Answer 2

then?

Answer 3

srry i gota do my problems bye i am sure smone else will help

Answer 4

wel i'll giv u hint

Answer 5

use s=ut+1/2 at^2 for horizontal component and get time substitute this time in any other eqn to get ur v

Answer 6

i get the answer 11.71m/s using formula s=vt-1/2at^2. is it correct?

Answer 7

how?

Answer 8

use equation v^2=u^2+2as

Answer 9

what is a?

Answer 10

a=-g(sin a)

Answer 11

\[-mg \sin \Theta =ma\]
sin theta=4.7

Answer 12

so our formula is matched but final a isn't

Answer 13

how could??

Answer 14

a=-g(sin a) & \[v=\sqrt{(v _{0}^{2}+2as)}\]

Answer 15

then,u get same answer as me or not?

Answer 16

i distracted no .never mind i'll find it

Answer 17

unbelievable..

Answer 18

i think 4.99 is correct hence particles velocity ought be decreased

Answer 19

okey. thanx a lot.

Answer 20

ur wellcome friend

Answer 21

no medals??