Question

imran can u help?

Answer 1

with?

Answer 2

physics projectile got some questions

Answer 3

sure shoot

Answer 4

wow thx dont go away let me type

Answer 5

1.If at any instant the velocity of a projectile be u and its direction of motion theta with horizontal then show that it will be moving at right angles to this direction after a time (u/g)*cosec theta

Answer 6

............

Answer 7

got it??

Answer 8

?pls eplain

Answer 9

h(t)= u cos(theta) t

v(t)= u sin(theta) t - 1/2 g t^2

Answer 10

?

Answer 11

??

Answer 12

what u did?

Answer 13

top one is horizontal component , bottom one is verticle -- that's why there gravitiy

Answer 14

but formula of height is 2u sintheta/g ryt??

Answer 15

where does that come from?

Answer 16

:) formula of max height

Answer 17

we are not loooking for max height

Answer 18

wel u can continue answering

Answer 19

k understood

Answer 20

..........

Answer 21

h(t)= u cos(theta) t

v(t)= u sin(theta) t - 1/2 g t^2


plug in (u/g)*cosec theta for t

in both equation

Answer 22

we are told to prove that time is u/g cosec theta

Answer 23

this is just verification isn't it???

Answer 24

we need to derive time=u/g cosec theta

Answer 25

We need to show that direction at that time is perpendicular to \[\theta\]

Answer 26

so?

Answer 27

k got it

Answer 28

what we get if we plug in u/g cosec theta??

Answer 29

some number which we can find direction by doing arctan{v/h}

Answer 30

how we find theta=90

Answer 31

tan 90 not defined

Answer 32

then how we take arc tan